Stopping potential for e- for wavelength 491 nm is 0.410 V incidence wavelength is changed to new value

Answer: The new wavelength is about 395 nm (shorter than 491 nm). None of the printed options matches; 645.83 nm cannot be correct.

Einstein's photoelectric equation:

\[eV_0 = \frac{hc}{\lambda} - \phi\]

Use \(hc = 1240\ \text{eV nm}\).

Step 1: work function.

\[\frac{hc}{\lambda_1} = \frac{1240}{491} = 2.525\ \text{eV}\]

\[\phi = 2.525 - 0.410 = 2.115\ \text{eV}\]

Step 2: new wavelength.

\[\frac{hc}{\lambda_2} = 1.02 + 2.115 = 3.135\ \text{eV}\]

\[\lambda_2 = \frac{1240}{3.135} \approx 395\ \text{nm}\]

Check: the stopping potential has increased, so each photon must carry more energy, i.e. the wavelength must be shorter than 491 nm. A wavelength of 645.83 nm gives photon energy 1.92 eV, which is less than the work function, so no photoelectrons would be emitted at all. The options seem to have been copied incorrectly.

Note: the earlier answer on this page says (d) 645.83 nm; the working above shows the new wavelength is about 395 nm.