Question

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

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

Answer 1

Answer: (D)

de-Broglie wavelength $\lambda$ for electron is given by:
$\lambda =(\frac{h}{2\times m\times K.E}{)}^{\frac{1}{2}}$ ….(i)
Where, $\lambda$ = wavelength of particle
$h$= planck’s constant
$K.E$ = kinetic energy of particle
Also, according to photoelectric effect: $K.E=hv-h{v}_0$
Where, $v$ = frequency of radiation and $v_0$ = threshold frequency
On substituting the value of K.E. in Eq...(i)
$\lambda =(\frac{h}{2\times m\times (v-v_0)}{)}^{\frac{1}{2}}$
= $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$