The work function of a metal is 4 eV. For the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3100 Å
By Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \). Photoelectrons of zero velocity have \( K_{\mathrm{max}}=0 \). So the energy of each photon just equals the work function, \( hc/\lambda=\phi_0 \).
Taking \( h=6.63\times10^{-34}\,\mathrm{J\,s} \), \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \):
\( \begin{aligned} \phi_0 &= 4\,\mathrm{eV} \\ &= 4\times1.6\times10^{-19}\,\mathrm{J} \\ &= 6.4\times10^{-19}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{hc}{\phi_0} \\ &= \frac{6.63\times10^{-34}\,\mathrm{J\,s}\times3\times10^{8}\,\mathrm{m\,s^{-1}}}{6.4\times10^{-19}\,\mathrm{J}} \\ &\approx 3.1\times10^{-7}\,\mathrm{m} \end{aligned} \)
Since 1 Å = \( 10^{-10}\,\mathrm{m} \), \( \lambda\approx3.1\times10^{-7}\,\mathrm{m} \) = 3100 Å.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy”
\( K_{\mathrm{max}}=h\nu-\phi_0 \qquad (11.2) \)
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Since Kmax must be non-negative, Eq. (11.2) implies that photoelectric emission is possible only if hν > φ0 or ν > ν0, where”
\( \displaystyle \nu_0=\frac{\phi_0}{h} \qquad (11.3) \)
“… Thus, there exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.”