Question 220: Dual Nature of Matter and Radiation

Question

Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $${{3\lambda } \over 4},$$ the speed of the fastest emitted electron will be:

Accepted Answer

$$h{v_0}^2 - h{v_0} = {1 \over 2}m{v^2}$$
$$	herefore$$ $${4 \over 3}h{v_0} - h{v_0} = {1 \over 2}mv{'^2}$$
$$	herefore$$ $${{v{'^2}} \over {{v^2}}} = {{{4 \over 3}v - {v_0}} \over {v - {v_0}}}$$
$$	herefore$$ $$v' = v\sqrt {{{{4 \over 3}v - {v_0}} \over {v - {v_0}}}} $$
$$	herefore$$ $$v' > v\sqrt {{4 \over 3}} $$

Answer

$$ = v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$

Answer

$$ = v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}$$

Answer

$$ > v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$

Answer

$$ < v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$

Dual Nature of Matter and Radiation

Radiation of wavelength $\lambda , is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to {{3\lambda } \over 4},$ the speed of the fastest emitted electron will be:

Choose the correct answer: