Atomic Structure - Result 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) $\frac{1}{2} \lambda$
Show Answer
Answer:
Correct Answer: 2. (a)
Solution:
- The expression of kinetic energy of photo electrons,
$$ KE=\frac{1}{2} m v^{2}=E-E _0 $$
When, $KE»E _0$, the equation becomes,
$$ \begin{gathered} KE=\frac{1}{2} m v^{2}=E \ \Rightarrow \quad \frac{1}{2} m v^{2}=\frac{h c}{\lambda} \Rightarrow \frac{p^{2}}{2 m^{2}}=\frac{h c}{\lambda} \ \Rightarrow \lambda=h c \times 2 m^{2} \times \frac{1}{p^{2}} \Rightarrow \lambda \propto \frac{1}{p^{2}} \ E=\frac{h c}{\lambda}=\text { energy of incident light. } \ E _0=\text { threshold energy or work functions, } \ \frac{1}{2} m v^{2}=\frac{1}{2} \times \frac{(m v)^{2}}{m^{2}}=\frac{1}{2} \times \frac{p^{2}}{m^{2}} \end{gathered} $$
$\because p=$ momentum $=m v$
As per the given condition,
$$ \frac{\lambda _2}{\lambda _1}=\left(\frac{p _1}{p _2}\right)^{2} $$
$$ \begin{array}{lll} \Rightarrow & \frac{\lambda _2}{\lambda}=\left(\frac{p}{1.5 \times p}\right)^{2}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9} \ \Rightarrow & \lambda _2=\frac{4}{9} \lambda & {\left[\begin{array}{c} \because \lambda _1=\lambda \ p _1=p \end{array}\right]} \end{array} $$
