exemplar-phy-12-dual-nature-of-radiation-and-matter

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Ques. Consider a beam of electrons (each electron with energy E$_{0}$) incident on a metal surface kept in an evacuated chamber. Then,
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(a) no electrons will be emitted as only photons can emit electrons
(b) electrons can be emitted but all with an energy, $E_0$
(c) electrons can be emitted with any energy, with a maximum of $E_0-\phi$ ( $\phi$ is the work function)
(d) electrons can be emitted with any energy, with a maximum of $E_0$
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From(i.), we note that if $V$ is inversely proportional to the wavelength $\lambda$. i.e., $V$ will increase with the decrease in the $\lambda$.
From Eq. (ii), we note that wavelength $\lambda$ is directly proportional to $\sin \theta$ and hence $\theta$.
So, with the decrease in $\lambda, \theta$ will also decrease.
Thus, when the voltage applied to $A$ is increased.
The diffracted beam will have the maximum at a value of $\theta$ that will be less than the earlier value.
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Ques. A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then, their de-Broglie wavelengths compare as
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(a) $\lambda_p=\lambda_n>\lambda_e>\lambda _{\alpha}$
(b) $\lambda _{\alpha}<\lambda_p=\lambda_n>\lambda_e$
(c) $\lambda_e<\lambda_p=\lambda_n>\lambda _{\alpha}$
(d) $\lambda_e=\lambda_p=\lambda_n=\lambda _{\alpha}$
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Thinking process :
The energy of any particle can be given by $K=\frac{1}{2} m v^{2}$ 
$\Rightarrow \quad m v=\sqrt{2 m k}$
Also, de-Broglie wavelength is given by
$\lambda=\frac{h}{m v}$
Now, relation between energy and wavelength of any particle is given by putting the value of Eq. (i) in Eq. (ii) 
$\lambda=\frac{h}{\sqrt{2 m k}}$
<Success>Sol. to be continued …</Success>
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$\therefore \lambda_p: \lambda_n: \lambda_e: \lambda _{\alpha}$
$\Rightarrow \frac{1}{\sqrt{m_p}}: \frac{1}{\sqrt{m_n}}: \frac{1}{\sqrt{m_e}}: \frac{1}{\sqrt{m _{\alpha}}}$
Since, $ m_p=m_n$, hence ${\lambda} _{p}=\lambda_n$
As, $ m _{\alpha}>m_p$, therefore $\lambda _{\alpha}<\lambda_p$
As, $ m_e<m_n$, therefore $\lambda_e>\lambda_n$
Hence, $ \lambda _{\alpha}<\lambda_p=\lambda_n<\lambda_e$
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Ques. An electron (mass $m$) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}(v_0>0)$ is in an electric field $\mathbf{E}=-E_0 \hat{\mathbf{i}}(E_0=$ constant $>0).$ It’s de-Broglie wavelength at time $t$ is given by
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(a) $\frac{\lambda_0}{\big(1+\frac{e E_0}{m} \frac{t}{v_0}\big)}$
(b) $\lambda_0 {\big(1+\frac{e E_0 t}{m v_0}\big)}$
(c) $\lambda_0$
(d) $\lambda_0 t $ $\newline $
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$\mathbf{F}=-e \mathbf{E}=-e[-E_0 \hat{\mathbf{i}}]=e E_0 \hat{\mathbf{i}}$
Acceleration of electron
$a=\frac{\mathbf{F}}{m}=\frac{e E_0 \hat{\mathbf{i}}}{m}$
Velocity of electron after time $t$,
$v =v_0 \hat{\mathbf{i}}+\frac{e E_0 \hat{\mathbf{i}}}{m} t=v_0+\frac{e E_0}{m} t \hat{\mathbf{i}} $
$ =v_0 1+\frac{e E_0}{m v_0} t \hat{\mathbf{i}} $
<Success>Sol. to be continued …</Success>
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Ques. An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ is in an electric field $\mathbf{E}=E_0 \hat{\mathbf{j}}$. If $\lambda_0=h / m v_0$, it’s de-Broglie wavelength at time $t$ is given by
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(a) $\lambda_0$
(b) $\lambda_0 \sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}}$
(c) $\frac{\lambda_0}{\sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}}}$
(d) $\frac{\lambda_0}{\big(1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}\big)}$
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$a=\frac{F}{m}=\frac{e E_0 \hat{j}}{m}$
It is acting along negative $y$-axis.
The initial velocity of electron along $x$-axis, $x_0=v_0 \hat{\mathbf{i}}$. Initial velocity of electron along $y$-axis,
$y_0=0 .$
Velocity of electron after time $t$ along $x$-axis, $x =_0 \hat{\mathbf{i}}$
Velocity of electron after timet along $y$-axis,
<Success>Sol. to be continued …</Success>
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$y=0+-\frac{e E_0}{m} \hat{{j}} t=-\frac{e E_0}{m} t \hat{{j}}$ 
Magnitude of velocity of electron after time $t$ is
$ v =\sqrt{v_x^{2}+v_y^{2}}=\sqrt{v_0^{2}+\frac{-e E_0}{m} t^{2}} $
$\Rightarrow =v_0 \sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}} $
de-Broglie wavelength, $\lambda^{\prime} =\frac{h}{m v} $
$ =\frac{h}{m v_0 \sqrt{1+e^{2} E_0^{2} t^{2} /(m^{2} v_0^{2})}} $
$ =\frac{\lambda_0}{\sqrt{1+e^{2} E_0^{2} t^{2} / m^{2} v_0^{2}}} $
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Ques. Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2} m v^{2}$, becomes comparable with $m c^{2}$, where $m$ is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?
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(a) $\lambda=10$ nm
(b) $\lambda=10^{-1}$ nm
(c) $\lambda=10^{-4}$ nm
(d) $\lambda=10^{-6}$ nm
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Thinking process :
The de-Broglie wavelength at which relativistic corrections become important must be greater than speed of light i.e., $3 \times 10^{8} m / s$.
Ans. (c, d)
de-Broglie wavelength
Here,
$ \lambda=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda} $
<Success>Sol. to be continued …</Success>
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(b) $\lambda_2=10^{-1} nm=10^{-1} \times 10^{-9} m=10^{-10} m$
$\Rightarrow \quad v_2=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-10}} \approx 10^{7} m / s$
(c) $\lambda_3=10^{-4} nm=10^{-4} \times 10^{-9} m=10^{-13} m$
$\Rightarrow \quad v_3=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-13}} \approx 10^{10} m / s$
(d) $\lambda_4=10^{-6} nm=10^{-6} \times 10^{-9} m=10^{-15} m$
$\Rightarrow \quad v_4=\frac{6.6 \times 10^{-34}}{9 \times 10^{-31} \times 10^{-15}} \approx 10^{12} m / s$
Thus, options (c) and (d) are correct as $v_3$ and $v_4$ is greater than $3 \times 10^{8} m / s$.
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Ques. Photons absorbed in matter are converted to heat. A source emitting $n$ photon/sec of frequency $v$ is used to convert 1 kg of ice at $0^{\circ} C$ to water at $0^{\circ} C$. Then, the time $T$ taken for the conversion
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(a) decreases with increasing $n$, with $v$ fixed
(b) decreases with $n$ fixed, $v$ increasing
(c) remains constant with $n$ and $v$ changing such that $n v=$ constant
(d) increases when the product $n v$ increases
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Ques. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values $\lambda_1, \lambda_2$ with $\lambda_1>\lambda_2$. Which of the following statement are true?
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(a) The particle could be moving in a circular orbit with origin as centre
(b) The particle could be moving in an elliptic orbit with origin as its focus
(c) When the de-Broglie wavelength is $\lambda_1$, the particle is nearer the origin than when its value is $\lambda_2$
(d) When the de-Broglie wavelength is $\lambda_2$, the particle is nearer the origin than when its value is $\lambda_1$
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Ans. (b, d)
The de-Broglie wavelength of the particle can be varying cyclically between two values $\lambda_1$ and $\lambda_2$, if particle is moving in an elliptical orbit with origin as its one focus.
Consider the figure given below
<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-193.jpg?height=188&width=381&top_left_y=365&top_left_x=620" height="160px">
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Let $v_1, v_2$ be the speed of particle at $A$ and $B$ respectively and origin is at focus $O$. If $\lambda_1, \lambda_2$ are the de-Broglie wavelengths associated with particle while moving at $A$ and $B$ respectively. Then, 
and
$ \lambda_1=\frac{h}{m v_1} $
$ \lambda_2=\frac{h}{m v_2} $
$ \therefore \frac{\lambda_1}{\lambda_2}=\frac{v_2}{v_1} $
$\text{ since } \lambda_1>\lambda_2 $
<Success>Sol. to be continued …</Success>
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Ques. (i) In the explanation of photoeletric effect, we assume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E _{\max }$ of the emitted electron as
$E _{\max }=h \nu-\phi_0$
where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $\nu$ ), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
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Ans.
(i) Here it is given that, an electron absorbs 2 photons each of frequency $v$ then $v^{\prime}=2 v$ where, $v^{\prime}$ is the frequency of emitted electron.
Given,
$ E _{\max }=h \nu-\phi_0$
Now, maximum energy for emitted electrons is
$E _{\max }^{\prime}=h(2 v)-\phi_0=2 h \nu-\phi_0$
(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible.
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Ans.
According to first statement, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is low when it has a longer wavelength.
But in second statement, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength.
This means in this statement material has to supply the energy for the emission of photons.
But this is not possible for a stable substances.
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Ans.
During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.
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and for the second condition,
Wavelength of light $\lambda^{\prime}=400 nm$
Also, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.
$\text{ i.e., } K _{\max }^{\prime} =2 K _{\max } $
$\text{ Here, } K _{\max }^{\prime} =\frac{h c}{\lambda}-\phi $
$\Rightarrow 2 K _{\max } =\frac{h c}{\lambda^{\prime}}-\phi_0 $
<Success>Sol. to be continued …</Success>
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Energy of falling photon of beam Energy of falling photon of beam
Now, according to question,
intensity of $A=$ intensity of $B$
$ \therefore \quad n_A h \nu_A=n_B h v_E $
$ \Rightarrow \quad \frac{v_A}{v_B}=\frac{n_B}{n_A}=\frac{n_B}{2 n_B}=\frac{1}{2} $
$ \Rightarrow \quad v_B=2 v_A $
Thus, from this relation we can infer that frequency of beam $B$ is twice of beam $A$.
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Ans.
 Given from conservation of momentum,
$|C|=|A|+|B| \frac{h}{\lambda_C}=\frac{h}{\lambda_A}+\frac{h}{\lambda_B} \quad \newline \because \lambda=\frac{h}{m v}=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda} \newline $
$\Rightarrow \frac{h}{\lambda_C}=\frac{h \lambda_B+h \lambda_A}{\lambda_A \lambda_B} \newline $
$\Rightarrow \frac{\lambda_C}{h}=\frac{\lambda_A \lambda_B}{h \lambda_A+h \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B} $
$\text{Case I}$ Suppose both $p_A$ and $p_B$ are positive, then
<Success>Sol. to be continued …</Success>
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$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
$\text{Case II}$ When both $p_A$ and $p_B$ are negative, then
$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
$\text{Case III}$ When $p_A>0, p_B<0$ i.e., $p_A$ is positive and $p_B$ is negative,
$ \frac{h}{\lambda_C}=\frac{h}{\lambda_A}-\frac{h}{\lambda_B}=\frac{(\lambda_B-\lambda_A) h}{\lambda_A \lambda_B} $
$ \Rightarrow \quad \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_B-\lambda_A} $
<Success>Sol. to be continued …</Success>
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Ans.
Given, $d=0.1 nm$,
Now, according to Bragg’s law
$\Rightarrow 2 d \sin \theta =n \lambda \Rightarrow 2 \times 0.1 \times \sin 30=1 \lambda \newline $
$\text{ Now, } \lambda =0.1 nm \Rightarrow=10^{-10} m \newline $
$\Rightarrow \lambda =\frac{h}{m v}=\frac{h}{p} \newline $
<Success>Sol. to be continued …</Success>
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We know that $6 \times 10^{26}$ atoms of Na weights $=23 kg$
So, volume of $6 \times 10^{6} Na$ atoms $=\frac{23}{0.97} m^{3}$
Volume occupied by one $Na$ atom $=\frac{23}{0.97 \times(6 \times 10^{26})}=3.95 \times 10^{-26} m^{3}$
Number of $Na$ atoms in target $({ }^{n} Na)$
$=\frac{10^{-7}}{3.95 \times 10^{-26}}=2.53 \times 10^{18}$
Let $n$ be the number of photons falling per second on the target.
<Success>Sol. to be continued …</Success>
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Energy of each photon $=h c / \lambda$
Total energy falling per second on target $=\frac{n h c}{\lambda}=I A$
$\therefore n =\frac{I A \lambda}{h c} $
$\Rightarrow \frac{100 \times 10^{-4} \times(660 \times 10^{-9})}{(6.62 \times 10^{-34}) \times(3 \times 10^{8})}=3.3 \times 10^{16} $
Let $P$ be the probability of emission per atom per photon.
The number of photoelectrons emitted per second
<Success>Sol. to be continued …</Success>
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$ 10^{-4} =P \times(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19}) $
$P =\frac{10^{-4}}{(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19})} $
$ =7.48 \times 10^{-21} $
$\therefore 10^{-4}=P \times(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19}) $
$\Rightarrow P=c $
Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.
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Thinking process :
$\text{ Work done by an external agency }=+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{1}{4} \int_0^{\infty} \frac{q^{2}}{x^{2}} d x$
Ans.
According to question, consider the figure given below From figure, $d=0.1 nm=10^{-10} m$,
$F=\frac{q^{2}}{4 \times 4 \pi \varepsilon_0 d^{2}}$
<Success>Sol. to be continued …</Success>
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$F_x=\frac{q^{2}}{4 \times 4 \pi \varepsilon_0 x^{2}}$
Work done by external agency in taking the electron from distance $d$ to infinity is
$W =\int_1^{\infty} F_x d x=\int_1^{\infty} \frac{q^{2} d x}{4 \times 4 \pi \varepsilon_0} \frac{1}{x^{2}} $
$ =\frac{q^{2}}{4 \times 4 \pi \varepsilon_0} \frac{1}{d} $
$ =\frac{(1.6 \times 10^{-19})^{2} \times 9 \times 10^{9}}{4 \times 10^{-10}} J $
$ =\frac{(1.6 \times 10^{-19})^{2} \times(9 \times 10^{9})}{(4 \times 10^{-10}) \times(1.6 \times 10^{-19})} eV=3.6 eV $
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Ques. Consider a 20 W bulb emitting light of wavelength 5000 $\AA$ and shining on a metal surface kept at a distance 2m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 $\AA$.
(i) Estimate number of photons emitted by the bulb per second.
[Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV) ?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?
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Ans.
Given, $P=20 W, \lambda=5000 \AA=5000 \times 10^{-10} m$
$d=2 m, \phi_0=2 eV, r=1.5 A=1.5 \times 10^{-10} m$
(i) Number of photon emitted by bulb per second is $n^{\prime}=\frac{p}{h c / \lambda}=\frac{p \lambda}{h c}$
$\Rightarrow$ $ =\frac{20 \times(5000 \times 10^{-10})}{(6.62 \times 10^{-34}) \times(3 \times 10^{8})} $
$ =5 \times 10^{19} s^{-1} $
<Success>Sol. to be continued …</Success>
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(ii) Energy of the incident photon $=\frac{h c}{\lambda}=\frac{(6.62 \times 10^{-34})(3 \times 10^{8})}{5000 \times 10^{-10} \times 1.6 \times 10^{-19}}$
$=2.48 eV$
As this energy is greater than $2 eV$ (i.e., work function of metal surface), hence photoelectric emission takes place.
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<Success>Sol. to be continued …</Success>
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(iii) Let $\Delta t$ be the time spent in getting the energy $\phi=$ (work function of metal).
Consider the figure,
$\Rightarrow \frac{P}{4 \pi d^{2}} \times \pi r^{2} \Delta t =\phi_0 $
$\Delta t =\frac{4 \phi_0 d^{2}}{P r^{2}} $
$ =\frac{4 \times(2 \times 1.6 \times 10^{-19}) \times 2^{2}}{20 \times(1.5 \times 10^{-10})^{2}} \approx 28.4 s $
(iv) Number of photons received by atomic disc in time $\Delta t$ is
<Success>Sol. to be continued …</Success>
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$ N =\frac{n^{\prime} \times \pi^{2}}{4 \pi d^{2}} \times \Delta t $
$ =\frac{n^{\prime} r^{2} \Delta t}{4 d^{2}} $
$ =\frac{(5 \times 10^{19}) \times(1.5 \times 10^{-10})^{2} \times 28.4}{4 \times(2)^{2}} \approx 2 $
(v) As time of emission of electrons is 11.04 s.
Hence, the photoelectric emission is not instantaneous in this problem.
In photoelectric emission, there is an collision between incident photon and free electron of the metal surface, which lasts for very very short interval of time $(\approx 10^{-9} s)$, hence we say photoelectric emission is instantaneous
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