Q195 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
When a light beam of frequency $\upsilon$ is incident on a metal surface, photoelectrons are emitted. If these electrons describe a circle of radius r in a magnetic field of strength B, then the work function of the metal is
$(h = $ Planck's constant, $m_e = $ mass of electron, $e = $ charge on electron$)$
$(h = $ Planck's constant, $m_e = $ mass of electron, $e = $ charge on electron$)$
- A$h\upsilon + \dfrac{Ber}{2m_e}$
- Correct answerB$h\upsilon - \dfrac{(Ber)^2}{2m_e}$
- C$h\upsilon + \dfrac{(Ber)^2}{2m_e}$
- D$h\upsilon - \dfrac{Ber}{2m_e}$
Solution
Using Einstein's photoelectric equation:
$K_{\max} = h\upsilon - W$
When an electron moves in a circular path in a uniform magnetic field, the necessary centripetal force is provided by the magnetic force:
$\dfrac{m_e v^2}{r} = e v B$
$v = \dfrac{e B r}{m_e}$
The maximum kinetic energy of the photoelectron is:
$K_{\max} = \dfrac{1}{2} m_e v^2 = \dfrac{1}{2} m_e \left( \dfrac{e B r}{m_e} \right)^2 = \dfrac{(B e r)^2}{2 m_e}$
Substituting $K_{\max}$ into the photoelectric equation:
$\dfrac{(B e r)^2}{2 m_e} = h\upsilon - W$
$W = h\upsilon - \dfrac{(B e r)^2}{2 m_e}$
Answer: $h\upsilon - \dfrac{(Ber)^2}{2m_e}$
$K_{\max} = h\upsilon - W$
When an electron moves in a circular path in a uniform magnetic field, the necessary centripetal force is provided by the magnetic force:
$\dfrac{m_e v^2}{r} = e v B$
$v = \dfrac{e B r}{m_e}$
The maximum kinetic energy of the photoelectron is:
$K_{\max} = \dfrac{1}{2} m_e v^2 = \dfrac{1}{2} m_e \left( \dfrac{e B r}{m_e} \right)^2 = \dfrac{(B e r)^2}{2 m_e}$
Substituting $K_{\max}$ into the photoelectric equation:
$\dfrac{(B e r)^2}{2 m_e} = h\upsilon - W$
$W = h\upsilon - \dfrac{(B e r)^2}{2 m_e}$
Answer: $h\upsilon - \dfrac{(Ber)^2}{2m_e}$
Hence, the correct answer is option: 'B'.