For a metal, the graph of stopping potential \(V_0\) against the frequency \(\nu\) of the incident light is a straight line. When this line is extended backwards, it meets the \(V_0\)-axis at \(-2.0\ \mathrm{V}\). Light of wavelength 248 nm falls on the metal. The maximum kinetic energy of the emitted photoelectrons is (Take \(hc=1240\ \mathrm{eV\,nm}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.0 eV
In terms of the stopping potential, Einstein’s equation is \( V_0=(h/e)\nu-\phi_0/e \). This is a straight line. At \( \nu=0 \) it gives \( V_0=-\phi_0/e \). So the point where the extended line meets the \( V_0 \)-axis gives the work function.
\( \begin{aligned} -\frac{\phi_0}{e} &= -2.0\ \mathrm{V} \\ \phi_0 &= 2.0\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} h\nu &= \frac{hc}{\lambda} \\ &= \frac{1240}{248} \\ &= 5.0\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} K_{\max} &= h\nu-\phi_0 \\ &= 5.0-2.0 \\ &= 3.0\ \mathrm{eV} \end{aligned} \)
Hence, the fastest photoelectrons have a kinetic energy of 3.0 eV.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… the photoelectric equation, Eq. (11.2), can be written as”
\( eV_0=h\nu-\phi_0;\ \text{for } \nu\ge\nu_0 \)
\( \text{or}\quad \)\( V_0=\left(\frac{h}{e}\right)\nu-\frac{\phi_0}{e} \)\( \qquad (11.4) \)
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”
\( K_{\max}=h\nu-\phi_0 \)\( \qquad (11.2) \)