Question

The threshold wavelength of a photosensitive material is equal to the frequency of H$_\alpha$ line of hydrogen. If a photon whose frequency equal to the frequency of H$_\beta$ line of hydrogen is incident on this photosensitive material, the maximum kinetic energy of the emitted photoelectrons is: (R -- Rydberg's constant, h -- Planck's constant and c -- speed of light in vacuum)

• A. \( \dfrac{Rhc}{1} \)
• B. \( \dfrac{5Rhc}{144} \)
• C. \( \dfrac{7Rhc}{144} \)
• D. \( \dfrac{Rhc}{36} \)

Hint:

For hydrogen spectral lines, use the Balmer series formula: \[ \nu = Rc \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \] and apply the photoelectric equation: \[ K.E._{\max} = h\nu_{\text{incident}} - h\nu_{\text{threshold}} \]

Answer 1

The Correct Option is C

Step 1: Use the photoelectric effect equation. \[ K.E._{\max} = h\nu_{\text{incident}} - h\nu_{\text{threshold}} \]
Step 2: Find the frequencies for H$_\alpha$ and H$_\beta$ lines (Balmer series). For H$_\alpha$ (\( n = 3 \to n = 2 \)): \[ \nu_\alpha = Rc \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = Rc \left( \frac{1}{4} - \frac{1}{9} \right) = Rc \left( \frac{5}{36} \right) \] For H$_\beta$ (\( n = 4 \to n = 2 \)): \[ \nu_\beta = Rc \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = Rc \left( \frac{1}{4} - \frac{1}{16} \right) = Rc \left( \frac{3}{16} \right) \]
Step 3: Substitute into the kinetic energy formula. \[ K.E._{\max} = h\nu_\beta - h\nu_\alpha = hRc \left( \frac{3}{16} - \frac{5}{36} \right) \] Find a common denominator: \[ \frac{3}{16} = \frac{27}{144}, \quad \frac{5}{36} = \frac{20}{144} \] \[ K.E._{\max} = hRc \left( \frac{27}{144} - \frac{20}{144} \right) = hRc \left( \frac{7}{144} \right) \] Final Answer: \[ \boxed{K.E._{\max} = \frac{7Rhc}{144}} \]