Atomic Structure Result Question 2
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) $\dfrac{4}{9} \lambda$
(b) $\dfrac{3}{4} \lambda$
(c) $\dfrac{2}{3} \lambda$
(d) $\dfrac{1}{2} \lambda$
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Answer:
Correct Answer: 2. ( a )
Solution:
- The expression of kinetic energy of photo electrons,
$ \mathrm{KE}=\dfrac{1}{2} m v^2=E-E_0 $
When, $\mathrm{KE}»E_0$, the equation becomes,
$ \begin{aligned} & \mathrm{KE}=\dfrac{1}{2} m v^2=E \\ & \Rightarrow \dfrac{1}{2} m v^2=\dfrac{h c}{\lambda} \Rightarrow \dfrac{p^2}{2 m^2}=\dfrac{h c}{\lambda} \\ & \Rightarrow \lambda=h c \times 2 m^2 \times \dfrac{1}{p^2} \Rightarrow \lambda \propto \dfrac{1}{p^2} \\ \end{aligned} $
$E=\dfrac{h c}{\lambda}=$ energy of incident light.
$E_0=$ threshold energy or work functions,
$ \dfrac{1}{2} m v^2=\dfrac{1}{2} \times \dfrac{(m v)^2}{m^2}=\dfrac{1}{2} \times \dfrac{p^2}{m^2} $
$\because p=$ momentum $=m v$
As per the given condition,
$ \dfrac{\lambda_2}{\lambda_1}=\left(\dfrac{p_1}{p_2}\right)^2 $
$ \begin{array}{lll} \Rightarrow & \dfrac{\lambda_2}{\lambda}=\left(\dfrac{p}{1.5 \times p}\right)^2=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9} & \\ \Rightarrow & \lambda_2=\dfrac{4}{9} \lambda & {\left[\begin{array}{c} \because \lambda_1=\lambda \\ p_1=p \end{array}\right]} \end{array} $
BETA
