Get Answers to all your Questions

header-bg qa

A small metal plate (work function \phi ) is kept at a distance d from a singly ionized, fixed ion. A monochromatic light beam is an incident on the metal plate. Find the minimum frequency of the light beam so that some of the photoelectrons may go around the ion in a circle.

Option: 1

\frac{1}{h}\left(\frac{e^2}{8 \pi \varepsilon_0 d}+e \phi\right)


Option: 2

\frac{1}{h}\left(\frac{e^2}{4 \pi \varepsilon_0 d}+e \phi\right)


Option: 3

\frac{1}{h}\left(\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{~d}}+\mathrm{e} \phi\right)


Option: 4

None of the Above


Answers (1)

best_answer

Suppose that the electrons are emitted with a velocity v, then
we can write,

\frac{1}{2} m v^2=h v-e \phi                     ...(1)

where \phi is the work function. If these electrons move in a circle of radius d,

\frac{e^2}{4 \pi \varepsilon_0 d^2}=\frac{m v^2}{d}                             ...(2)

\text { or, } \quad \mathrm{mv}^2=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{~d}}

\text { Thus, } \quad h v=\frac{e^2}{8 \pi \varepsilon_0 d}+e \phi

\text { or, } v=\frac{1}{h}\left(\frac{e^2}{8 \pi \varepsilon_0 d}+e \phi\right)

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE