Q146 / 199 · NEET 2008 (Mains) · Physics

The work function of a surface of a photosensitive material is $6.2 \mathrm{eV}$. The wavelength of the incident radiation for which the stopping potential is $5 \mathrm{~V}$ lies in the

  • A
    X-ray region
  • Correct answer
    B
    Ultraviolet region
  • C
    Visible region
  • D
    Infrared region

Solution

$$ \begin{aligned} & \mathrm{eV}_0=E-\phi \\ & E=\mathrm{eV}_0+\phi \\ & =5 \mathrm{eV}+6.2 \mathrm{eV} \\ & =11.2 \mathrm{eV} \\ & \therefore \quad \lambda=\left(\frac{12400}{11.2}\right) Å=1000 Å \\ & \Rightarrow \text { hence lies in ultraviolet region. } \\ & \end{aligned} $$

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

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