Q155 / 160 · AP EAMCET 2023 (15 May Shift 2) · Physics

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$, when light frequency $v$ incidents on it is $\lambda$. Then $\mathrm{v}=$ ( $\mathrm{h}$ - Planck's constant, $\mathrm{m}$ - mass of electron)

  • A
    $\frac{2 \phi}{\mathrm{h}}-\frac{\mathrm{h}}{\mathrm{m} \lambda^2}$
  • B
    $\frac{2 \phi}{\mathrm{h}}+\frac{\mathrm{h}}{\mathrm{m} \lambda^2}$
  • Correct answer
    C
    $\frac{\phi}{h}+\frac{h}{2 m \lambda^2}$
  • D
    $\frac{\phi}{\mathrm{h}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda^2}$

Solution

By photoelectric effect $\begin{aligned} & \mathrm{hv}=\phi+\mathrm{K} \cdot \mathrm{E} \\ & \mathrm{h} v=\phi+\frac{1}{2} \mathrm{mv}^2 \\ & v=\frac{\phi}{\mathrm{h}}+\frac{\mathrm{mv}^2}{2 \mathrm{~h}} \\ & v=\frac{\phi}{\mathrm{h}}+\frac{\mathrm{m}}{2 \mathrm{~h}} \times \frac{\mathrm{h}^2}{\mathrm{~m}^2 \lambda^2} \\ & \because \lambda=\frac{\mathrm{h}}{\mathrm{mv}} ; \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda} \\ & \mathrm{V}=\frac{\phi}{\mathrm{h}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda^2} \end{aligned}$

Hence, the correct answer is option: 'C'.

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