Modern Physics 2 Question 34
33. Photoelectrons are emitted when $400 nm$ radiation is incident on a surface of work function $1.9 eV$. These photoelectrons pass through a region containing $\alpha$-particles. A maximum energy electron combines with an $\alpha$-particle to form a $He^{+}$ ion, emitting a single photon in this process. $He^{+}$ions thus formed are in their fourth excited state. Find the energies in $eV$ of the photons lying in the 2 to $4 eV$ range, that are likely to be emitted during and after the combination. (1999, 5M) [Take $h=4.14 \times 10^{-15} eV$-s]
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Answer:
Correct Answer: 33. During combination $3.4 eV$. After combination $3.84 eV, 2.64 eV$ 34. (a) $10 % / s$ (b) 285.1 (d) $111 s$
Solution:
- Given work function, $W=1.9 eV$
Wavelength of incident light, $\lambda=400 nm$
$\therefore$ Energy of incident light, $E=\frac{h c}{\lambda}=3.1 eV$
(Substituting the values of $h, c$ and $\lambda$ )
Therefore, maximum kinetic energy of photoelectrons
$$ K _{\max }=E-W=(3.1-1.9)=1.2 eV $$
Now the situation is as shown below :
Energy of electron in $4^{\text {th }}$ excited state of $He^{+}(n=5)$ will be
$E _5=-13.6 \frac{Z^{2}}{n^{2}} eV \Rightarrow E _5=-(13.6) \frac{(2)^{2}}{(5)^{2}}=-2.2 eV$
Therefore, energy released during the combination
$$ =1.2-(-2.1)=3.4 eV $$
Similarly, energies in other energy states of $He^{+}$will be
$$ \begin{aligned} & E _4=-13.6 \frac{(2)^{2}}{(4)^{2}}=-3.4 eV \\ & E _3=-13.6 \frac{(2)^{2}}{(3)^{2}}=-6.04 eV \\ & E _2=-13.6 \frac{(2)^{2}}{2^{2}}=-13.6 eV \end{aligned} $$
The possible transitions are
$$ \begin{aligned} \Delta E _{5 \rightarrow 4} & =E _5-E _4=1.2 eV<2 eV \\ \Delta E _{5 \rightarrow 3} & =E _5-E _3=3.84 eV \\ \Delta E _{5 \rightarrow 2} & =E _5-E _2=11.4 eV>4 eV \\ \Delta E _{4 \rightarrow 3} & =E _4-E _3=2.64 eV \\ \Delta E _{4 \rightarrow 2} & =E _4-E _2=10.2 eV>4 eV \end{aligned} $$
Hence, the energy of emitted photons in the range of $2 eV$ and $4 eV$ are
$3.4 eV$ during combination and
$3.84 eV$ and 2.64 after combination.
