If the maximum kinetic energy of emitted elec | vedclass

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

Question

(D) $1$. The stopping potential $V_s$ is related to the maximum kinetic energy $(K.E.)_{\max}$ by the equation: $(K.E.)_{\max} = e V_s$.Given $(K.E.)_

Vedclass · Accepted Answer

$2 	ext{ V}, 3000 	ext{ Å}$

Vedclass · Answer

(D) $1$. The stopping potential $V_s$ is related to the maximum kinetic energy $(K.E.)_{\max}$ by the equation: $(K.E.)_{\max} = e V_s$.Given $(K.E.)_{\max} = 3.2 	imes 10^{-19} 	ext{ J}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$.$V_s = \frac{(K.E.)_{\max}}{e} = \frac{3.2 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 2 	ext{ V}$.$2$. The threshold wavelength $\lambda_0$ is related to the work function $\Phi$ by the equation: $\Phi = \frac{hc}{\lambda_0}$.Given $\Phi = 6.63 	imes 10^{-19} 	ext{ J}$,$h = 6.63 	imes 10^{-34} 	ext{ J} \cdot 	ext{s}$,and $c = 3 	imes 10^{8} 	ext{ m/s}$.$\lambda_0 = \frac{hc}{\Phi} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{6.63 	imes 10^{-19}} = 3 	imes 10^{-7} 	ext{ m} = 3000 	imes 10^{-10} 	ext{ m} = 3000 	ext{ Å}$.Thus,the stopping potential is $2 	ext{ V}$ and the threshold wavelength is $3000 	ext{ Å}$. The correct option is $(D)$.

Similar Questions

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)

In a photoelectric experiment,the wavelength of the light incident on a metal is changed from $300\, nm$ to $400\, nm$. The decrease in the stopping potential is close to ................ $V$ $\left( \frac{hc}{e} = 1240\, nm \cdot V \right)$

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

Vedclass Test Series

Exam Paper Generator

Online Exam Module