MCQ
The de Broglie wavelength $\left( \lambda  \right)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, [$v_0$ is threshold frequency]:
  • A
    $\lambda  \propto \frac{1}{{\left( {v - {v_0}} \right)}}$
  • B
    $\lambda  \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{4}}}}}$
  • C
    $\lambda  \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{3}{2}}}}}$
  • $\lambda  \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{2}}}}}$

Answer

Correct option: D.
$\lambda  \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{2}}}}}$
d
$\lambda  = \frac{h}{{mv}}$

According to Einstein's theory of photoelectric effect:

$hv = h{v_0} + KE$

$hv = h{v_0} + \frac{1}{2}m{v^2}$

$2h(v - {v_0}) = m{v^2}$

$\frac{{2h(v - {v_0})}}{m} = {v^2}$

$v \propto {(v - {v_0})^{\frac{1}{2}}}$

$\lambda  \propto \frac{h}{{m{{(v - {v_0})}^{\frac{1}{2}}}}}$

$\lambda  \propto \frac{1}{{{{(v - {v_0})}^{\frac{1}{2}}}}}$

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