Light of frequency \( 8.27 \times 10^{14}\,\mathrm{Hz} \) is incident on a metal surface. Electrons with a maximum speed of \( 6.6 \times 10^{5}\,\mathrm{m/s} \) are ejected from the surface. The threshold frequency for photoemission of electrons will be
(Given: h = \( 6.6 \times 10^{-34}\,\mathrm{J\,s} \) and mass of electron = \( 9 \times 10^{-31}\,\mathrm{kg} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.30 \times 10^{14}\,\mathrm{Hz} \)
Einstein's photoelectric equation gives the maximum kinetic energy of the ejected electrons:
\( \displaystyle h\nu = h\nu_0 + K_{\max} \;\Rightarrow\; \nu_0 = \nu - \frac{K_{\max}}{h} \)
\( \displaystyle K_{\max} = \frac{1}{2}mv_{\max}^{2} = \frac{1}{2} \times 9\times10^{-31} \times (6.6\times10^{5})^{2} = 1.96\times10^{-19}\,\mathrm{J} \)
\( \displaystyle \frac{K_{\max}}{h} = \frac{1.96\times10^{-19}}{6.6\times10^{-34}} = 2.97\times10^{14}\,\mathrm{Hz} \)
\( \displaystyle \nu_0 = 8.27\times10^{14} - 2.97\times10^{14} = 5.30\times10^{14}\,\mathrm{Hz} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
