Question

Two sources P and Q produce electromagnetic waves of wavelengths $\lambda$ and $2\lambda$, respectively. Source P ejects a photon with maximum kinetic energy 4.0 eV from a metal with work function 2.0 eV. The maximum kinetic energy (eV) of a photon ejected by source Q from the same metal is .............

Hint:

If the photon energy is less than the work function, no photoelectron is emitted → KE = 0 eV.

Answer 1

Correct Answer: 1

Step 1: Use photoelectric equation.
\[ K_{\max} = h\nu - \phi \] For source P: \[ K = 4\ \text{eV},\ \phi = 2\ \text{eV} \] \[ h\nu_P = 6\ \text{eV} \] Step 2: Find energy of Q.
Wavelength doubled → frequency halved: \[ \nu_Q = \frac{\nu_P}{2} \] Thus photon energy: \[ h\nu_Q = 3\ \text{eV} \] Step 3: Calculate KE for source Q.
\[ K_Q = h\nu_Q - \phi = 3 - 2 = 1\ \text{eV} \] But 3 eV is not enough to eject a photoelectron beyond the work function threshold (minimum required = 2 eV). Thus maximum KE = 0 eV.
Step 4: Conclusion.
Source Q fails to produce photoelectrons with positive KE → answer = 0 eV.