The threshold wavelength for a metal is 495 nm. The work function of the metal is (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\), \(c=3\times10^{8}\ \mathrm{m\,s^{-1}}\) and \(1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2.5 eV
The work function is the least energy an electron needs to escape from the metal surface. At the threshold wavelength, each photon has just this energy, so \( \phi_0=hc/\lambda_0 \). The result comes out in joules. An electron gains 1 eV of energy across a potential difference of 1 V. So dividing the energy in joules by \( 1.6\times10^{-19} \) gives it in eV.
Given:
\( \lambda_0=495\ \mathrm{nm}=495\times10^{-9}\ \mathrm{m} \)
\( \begin{aligned} \phi_0 &= \frac{hc}{\lambda_0} \\ &= \frac{6.6\times10^{-34}\times3\times10^{8}}{495\times10^{-9}} \\ &= 4.0\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} \phi_0 &= \frac{4.0\times10^{-19}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &= 2.5\ \mathrm{eV} \end{aligned} \)
Therefore, the work function of the metal is 2.5 eV.
NCERT (2026-27):
Example 11.2 – “(a) For the cut-off or threshold frequency, the energy hν0 of the incident radiation must be equal to work function φ0 …”
Electron Emission – “… One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt, so that 1 eV = 1.602 ×10–19 J.”