Question 175: Dual Nature of Matter and Radiation

Question

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

Accepted Answer

From the photoelectric effect equation
$${{hc} \over \lambda } = \phi  + e{v_s}$$
so, $$e{v_{{s_1}}} = {{hc} \over {{\lambda _1}}} - \phi $$ .....(i)
$$e{v_{{s_2}}} = {{hc} \over {{\lambda _2}}} - \phi $$ ......(ii)
Subtract equation (i) from equation (ii)
$$e{v_{{s_1}}} - e{v_{{s_2}}} = {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _2}}}$$
$${v_{{s_1}}} - {v_{{s_2}}} = {{hc} \over e}\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$$
$$(0.710 - 1.43) = 1240\left( {{1 \over {491}} - {1 \over {{\lambda _2}}}} \right)$$
$${{ - 0.72} \over {1240}} = {1 \over {491}} - {1 \over {{\lambda _2}}}$$
$${1 \over {{\lambda _2}}} = {1 \over {491}} + {{0.72} \over {1240}}$$
$${1 \over {{\lambda _2}}} = 0.00203 + 0.00058$$
$${1 \over {{\lambda _2}}} = 0.00261$$
$${\lambda _2} = 383.14$$
$${\lambda _2} \simeq 382$$ nm

Answer

400 nm

Answer

329 nm

Answer

309 nm

Answer

382 nm

Dual Nature of Matter and Radiation

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

Choose the correct answer: