Question:
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\()\)
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\()\)
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The magnetic force gives the speed of the electron from the radius, and the photoelectric equation then gives the work function.
Updated On: Nov 12, 2026
- \(h\upsilon +\frac{Ber}{2m_{e}}\)
- \(h\upsilon -\frac{(Ber)^{2}}{2m_{e}}\)
- \(h\upsilon +\frac{(Ber)^{2}}{2m_{e}}\)
- \(h\upsilon -\frac{Ber}{2m_{e}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
In a magnetic field the electron moves in a circle, so the magnetic force provides the centripetal force: \(Bev = \dfrac{m_e v^2}{r}\).
Step 2: Kinetic energy
This gives \(v = \dfrac{Ber}{m_e}\), so \(K_{max} = \dfrac{1}{2}m_e v^2 = \dfrac{(Ber)^2}{2m_e}\).
Step 3: Photoelectric equation
\(h\upsilon = \phi + K_{max}\), so \(\phi = h\upsilon - \dfrac{(Ber)^2}{2m_e}\).
Step 4: Compare the options
The work function is less than \(h\upsilon\) and has the squared term \(\dfrac{(Ber)^2}{2m_e}\), which matches the second option.
Final Answer:
The work function is \(h\upsilon\) minus \((Ber)^2/2m_e\). \[ \boxed{\text{(B) }h\upsilon -\dfrac{(Ber)^{2}}{2m_{e}}} \]
In a magnetic field the electron moves in a circle, so the magnetic force provides the centripetal force: \(Bev = \dfrac{m_e v^2}{r}\).
Step 2: Kinetic energy
This gives \(v = \dfrac{Ber}{m_e}\), so \(K_{max} = \dfrac{1}{2}m_e v^2 = \dfrac{(Ber)^2}{2m_e}\).
Step 3: Photoelectric equation
\(h\upsilon = \phi + K_{max}\), so \(\phi = h\upsilon - \dfrac{(Ber)^2}{2m_e}\).
Step 4: Compare the options
The work function is less than \(h\upsilon\) and has the squared term \(\dfrac{(Ber)^2}{2m_e}\), which matches the second option.
Final Answer:
The work function is \(h\upsilon\) minus \((Ber)^2/2m_e\). \[ \boxed{\text{(B) }h\upsilon -\dfrac{(Ber)^{2}}{2m_{e}}} \]
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