Question

The de Broglie wavelength $\left(\lambda \right)$ associated with a photoelectron varies with the frequency $\left(v\right)$ of the incident radiation as, $[v_0$ is threshold frequency]:

• A. $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{3}{2}}}$
• B. $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$
• C. $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{4}}}$
• D. $\lambda \propto \frac{1}{(v-v_0)}$

Answer 1

The correct option is B $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$

In photoelectric effect, incident energy = thershold energy + KE
$hv=h{v}_0+\text{KE}$
$\text{KE}=hv-h{v}_0$
KE = $\frac{m{v}^2}{2}=h(v-v_0)$
v = $\sqrt{\frac{2h(v-v_0)}{m}}$
de broglie wavelength $\lambda$ = $\frac{h}{mv}$
v = $\frac{h}{m\lambda }$. Substituting v,
v = $\frac{h}{m\lambda }$.= $\sqrt{\frac{2h(v-v_0)}{m}}$
or $\lambda$ = $\frac{h}{m}X\sqrt{\frac{m}{2h(v-v_0)}}$
$\lambda$ = $\sqrt{\frac{h}{2m(v-v_0)}}$
$\lambda$ = $(\frac{h}{2m(v-v_0)}{)}^{1/2}$
Since h and m are constants,
$\lambda$ $\propto$ $(\frac{1}{(v-v_0)}{)}^{1/2}$
Hence, the correct option is option (b).