Atomic Structure 1 Question 26
2. If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)
(2019 Main, 8 April II)
(a) $\frac{4}{9} \lambda$
(b) $\frac{3}{4} \lambda$
(c) $\frac{2}{3} \lambda$
(d) $\frac{1}{2} \lambda$
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Answer:
Correct Answer: 2. (a)
Solution:
- The expression of kinetic energy of photo electrons,
$$ \mathrm{KE}=\frac{1}{2} m v^{2}=E-E_{0} $$
When, $\mathrm{KE}»E_{0}$, the equation becomes,
$$ \begin{gathered} \mathrm{KE}=\frac{1}{2} m v^{2}=E \ \Rightarrow \quad \frac{1}{2} m v^{2}=\frac{h c}{\lambda} \Rightarrow \frac{p^{2}}{2 m^{2}}=\frac{h c}{\lambda} \ \Rightarrow \lambda=h c \times 2 m^{2} \times \frac{1}{p^{2}} \Rightarrow \lambda \propto \frac{1}{p^{2}} \ E=\frac{h c}{\lambda}=\text { energy of incident light. } \ E_{0}=\text { threshold energy or work functions, } \ \frac{1}{2} m v^{2}=\frac{1}{2} \times \frac{(m v)^{2}}{m^{2}}=\frac{1}{2} \times \frac{p^{2}}{m^{2}} \end{gathered} $$
$\because p=$ momentum $=m v$
As per the given condition,
$$ \frac{\lambda_{2}}{\lambda_{1}}=\left(\frac{p_{1}}{p_{2}}\right)^{2} $$
$$ \begin{array}{lll} \Rightarrow & \frac{\lambda_{2}}{\lambda}=\left(\frac{p}{1.5 \times p}\right)^{2}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9} \ \Rightarrow & \lambda_{2}=\frac{4}{9} \lambda & {\left[\begin{array}{c} \because \lambda_{1}=\lambda \ p_{1}=p \end{array}\right]} \end{array} $$
