Question

Suppose a photon of frequency (v) causes photoelectric emission from a surface with threshold frequency $\left(v_o\right)$, what is the de-Broglie wavelength of the photoelectron of maximum kinetic energy in terms of $\Delta v=v-v_0$ ?

• A. $\Delta v=\frac{h}{Zm\lambda }$
• B. $\Delta v=\frac{h}{\lambda }$
• C. $[\frac{1}{v_0}-\frac{1}{v}]=\frac{m{c}^2}{h}$
• D. $\lambda ={\left(\frac{h}{2\pi \Delta v}\right)}^{1/2}$

Answer 1

Answer: (A)

$\Delta v=\frac{h}{Zm\lambda }$

$hv=h{v}_0+K.E.$
$K.E=\frac{p^2}{2m}$
Also A to de broglie $e{q}^n\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$
$KE=\frac{(h/\lambda {)}^2}{2m}=\frac{h^2}{{\lambda }^{2}2m}$
Put in equation,
$hv=h{v}_0+\frac{h^2}{{\lambda }^{2}2m}$
$h(v-v_0)=\frac{h^2}{{\lambda }^{2}2m}$
${\lambda }^2=\frac{h}{2m(v-v_0)}=\frac{h}{2m(v-v_0)}$
$\lambda ={\left[\frac{h}{2m(v-v_0)}\right]}^{\frac{1}{2}}={\left[\frac{h}{2m\Delta v}\right]}^{\frac{1}{2}}$.