An electron in the hydrogen atom jumps from e | vedclass

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

Question

(C) Given,stopping potential $V_{0} = 10 \ V$ and work function $W = 2.75 \ eV$.According to Einstein's photoelectric equation,the energy of the incid

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given,stopping potential $V_{0} = 10 \ V$ and work function $W = 2.75 \ eV$.According to Einstein's photoelectric equation,the energy of the incident photon is $E = h
u = eV_{0} + W$.Substituting the values,$E = 10 \ eV + 2.75 \ eV = 12.75 \ eV$ ..... $(i)$.For a hydrogen atom,the energy of a photon emitted during a transition from state $n$ to the ground state $(n=1)$ is given by $h
u = E_{n} - E_{1}$.Since $E_{n} = -\frac{13.6}{n^{2}} \ eV$,we have $h
u = -\frac{13.6}{n^{2}} - (-13.6) = 13.6 \left(1 - \frac{1}{n^{2}}ight) \ eV$.Equating this to the energy from $(i)$,$13.6 \left(1 - \frac{1}{n^{2}}ight) = 12.75$.$1 - \frac{1}{n^{2}} = \frac{12.75}{13.6} = 0.9375$.$\frac{1}{n^{2}} = 1 - 0.9375 = 0.0625$.$n^{2} = \frac{1}{0.0625} = 16$.Therefore,$n = 4$.

An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.75 \ eV$. If the stopping potential of the photoelectron is $10 \ V$,then the value of $n$ is

Similar Questions

The variation of photo-current with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph. Then:

In the case of photoelectric emission from a certain metal,the cutoff frequency is $v$. If radiation of frequency $2v$ is incident on the metal plate,the maximum possible velocity of the emitted electrons will be ($m=$ mass of electron).

The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy $E$. If $E$ is to be increased to twice the original value,the wavelength changes to $\lambda_1$. Which of the following is correct?

In photoelectric effect, the graph of stopping potential $(V_0)$ versus frequency $(\nu)$ is a straight line. The slope of this graph is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module