Monochromatic light of wavelength lambda is i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) As photoelectrons are emitted,the sphere becomes positively charged. The emission stops when the potential $V$ of the sphere is such that the maxi

Vedclass · Accepted Answer

$\frac{4\pi \epsilon_0 ahc}{e^2} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$

Vedclass · Answer

(C) As photoelectrons are emitted,the sphere becomes positively charged. The emission stops when the potential $V$ of the sphere is such that the maximum kinetic energy of the emitted photoelectrons is equal to the work done against the electrostatic potential.According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$.The emission stops when $eV = K_{max}$,where $V$ is the potential of the sphere.Thus,$V = \frac{hc}{e} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.The charge $Q$ on the sphere of radius $a$ is given by $Q = CV = (4\pi \epsilon_0 a) V$.Substituting $V$: $Q = 4\pi \epsilon_0 a \cdot \frac{hc}{e} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.The number of photoelectrons emitted $n$ is given by $n = \frac{Q}{e} = \frac{4\pi \epsilon_0 ahc}{e^2} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.

Monochromatic light of wavelength $\lambda$ is incident on an isolated metal sphere of radius $a$. The threshold wavelength is $\lambda_0$,which is greater than $\lambda$. How many photoelectrons will be emitted before the emission of photoelectrons stops?

Similar Questions

For intensity $I$ of a light of wavelength $5000 \, Å$, the photoelectron saturation current is $0.40 \, μA$ and the stopping potential is $1.36 \, V$. The work function of the metal is ........... $eV$.

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

If the frequency of incident light falling on a metallic surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

The following graphs show the variation of stopping potential $(V_s)$ corresponding to the frequency of incident radiation $(f)$ for a given metal. The correct variation is shown in graph ($f_0 =$ Threshold frequency):

Vedclass Test Series

Exam Paper Generator

Online Exam Module