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\()\)

• A. \(h\upsilon +\frac{Ber}{2m_{e}}\)
• B. \(h\upsilon -\frac{(Ber)^{2}}{2m_{e}}\)
• C. \(h\upsilon +\frac{(Ber)^{2}}{2m_{e}}\)
• D. \(h\upsilon -\frac{Ber}{2m_{e}}\)

Hint:

The magnetic force gives the speed of the electron from the radius, and the photoelectric equation then gives the work function.

Answer 1

The Correct Option is B

Step 1: Dimension check
A kinetic energy must have a squared speed, so the correction term has to contain $(Ber)^2$, which removes the options with $Ber$ to the first power.

Step 2: Sign
The photon energy is shared between the work function and the kinetic energy, so $\phi = h\upsilon - K_{max}$. The sign must be negative.

Step 3: Conclusion
Together these leave $h\upsilon - \dfrac{(Ber)^2}{2m_e}$.

Final Answer:
Only the second option has both the squared term and the minus sign. \[ \boxed{\text{(B) }h\upsilon -\dfrac{(Ber)^{2}}{2m_{e}}} \]