The surface of a metal is first illuminated with a light of wavelength 300 nm and later illuminated by…

The surface of a metal is first illuminated with a light of wavelength 300 nm and later illuminated by another light of wavelength 500 nm . It is observed that the ratio of maximum velocities of photo electrons in two cases is 3. The work function of metal value is close to
  1. 6.48 eV
  2. 1.23 eV
  3. 4.17 eV
  4. 2.28 eV

Solution

$\lambda_1=300 \mathrm{~nm}, \lambda_2=500 \mathrm{~nm}, \frac{\mathrm{v}_1}{\mathrm{v}_2}=3$
By Einstein photoelectric equation, \(\mathrm{E}=\phi_0+\frac{1}{2} \mathrm{mv}^2\) \(\Rightarrow\left(\frac{\mathrm{v}_1}{\mathrm{v}_2}\right)^2=\frac{\mathrm{E}_1-\phi_0}{\mathrm{E}_2-\phi_0}=\frac{\left(\frac{1240}{\lambda_1}-\phi_0\right)}{\left(\frac{1240}{\lambda_2}-\phi_0\right)}\) \(\Rightarrow 9=\frac{\left(\frac{1240}{300}-\phi_0\right)}{\left(\frac{1240}{500}-\phi_0\right)} \Rightarrow 9\left(\frac{124}{50}-\phi_0\right)=\left(\frac{124}{30}-\phi_0\right)\) \(\therefore \phi_0 \approx 2.28 \mathrm{eV}\)

Asked in: AP EAMCET 2024 (22 May Shift 1)

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