Modern Physics 2 Question 9
9. Surface of certain metal is first illuminated with light of wavelength $\lambda _1=350 nm$ and then by light of wavelength $\lambda _2=540 n-m$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2 . The work function of the metal (in eV) is close to (energy of photon $=\frac{1240}{\lambda(\text { in } n-m)} eV$ )
(a) 5.6
(b) 2.5
(c) 1.8
(d) 1.4
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Solution:
- Let maximum speed of photo electrons in first case is $v _1$ and maximum speed of photo electrons in second case is $v _2$
Assumption I if we assume difference in maximum speed in two cases is 2 then $v _1=v$ and $v _2=3 v$
According to Einstein’s photo electron equation Energy of incident photon $=$ work function $+KE$
$$ \text { i.e. } \quad \frac{h c}{\lambda}=\varphi _0+\frac{1}{2} m v^{2} $$
Where $h c=1240 eV, \lambda$ is wavelength of light incident, $\varphi _0$ is work function and $v$ is speed of photo electrons.
$$ \begin{aligned} & \text { When } & \lambda _1 & =350 nm \\ & \therefore & \frac{h c}{350} & =\varphi _0+\frac{1}{2} m v^{2} \\ & \text { or } & \frac{h c}{350}-\varphi _0 & =\frac{1}{2} m v^{2} \\ & \text { when } & \lambda _2 & =540 nm \\ & \therefore & \frac{h c}{540} & =\varphi _0+\frac{1}{2} m\left(3 v^{2}\right) \\ & \therefore & \frac{h c}{540}-\varphi _0 & =\left(\frac{1}{2} m v^{2}\right) \times 9 \end{aligned} $$
Now, we divide Eq. (i) by Eq. (ii), we get
$$ \begin{array}{rlrl} \frac{\frac{h c}{350}-\varphi _0}{\frac{h c}{540}-\varphi _0} & =\frac{\frac{1}{2} m v^{2}}{\left(\frac{1}{2} m v^{2}\right) \times 9}=\frac{1}{9} \\ \text { or } 9 \frac{h c}{350}-\varphi _0 & =\frac{h c}{540}-\varphi _0 \\ \text { or } \quad & 8 \varphi _0 & =h c \frac{9}{350}-\frac{1}{540} \\ \text { or } & & \varphi _0 & =\frac{1}{8} \times 1240 \frac{9 \times 540-350}{350 \times 540}=3.7 eV . \end{array} $$
No option given is correct.
Alternate Method
Assumption II If we assume velocity of one is twice in factor with second, then.
Let $v _1=2 v$ and $v _2=v$
We know that from Einstein’s photoelectric equation, energy of incident radiation $=$ work function $+KE$
or $\quad \frac{h c}{\lambda}=\varphi+\frac{1}{2} m v^{2}$
Let when $\lambda _1=350 nm$ then $v _1=2 v$
and when $\lambda _1=540 nm$ then $v _2=v$
$\therefore$ Above Eq. becomes
$$ \begin{aligned} & \frac{h c}{\lambda _1} & =\varphi+\frac{1}{2} m\left(2 v^{2}\right) \\ \text { or } & \frac{h c}{\lambda _1}-\varphi & =\frac{1}{2} m \times 4 v^{2} \\ \text { and } & \frac{h c}{\lambda _2} & =\varphi+\frac{1}{2} m v^{2} \\ \text { or } & \frac{h c}{\lambda _2}-\varphi & =\frac{1}{2} m v^{2} \end{aligned} $$
Now, we divide Eq. (i) by (ii) Eq.
or
$$ \frac{\frac{h c}{\lambda _1}-\varphi}{\frac{h c}{\lambda _2}-\varphi}=\frac{\frac{1}{2} m \times 4 v^{2}}{\frac{1}{2} m v^{2}}=4 $$
or
$$ \frac{h c}{\lambda _1}-\varphi=\frac{4 h c}{\lambda _2}-4 \varphi $$
$$ \begin{aligned} \text { or } \varphi=\frac{1}{3} h c \frac{4}{\lambda _2}-\frac{1}{\lambda _1} & =\frac{1}{3} \times 1240 \frac{4 \times 350-540}{350 \times 540} \\ \text { or } \quad \varphi & =1.8 eV \end{aligned} $$
According the assumption II, correct option is (c).
