A metallic surface is illuminated with radiat | vedclass

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

Question

(D) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} -

Vedclass · Accepted Answer

$3 \lambda$

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$When the wavelength is changed to $2\lambda$,the stopping potential becomes $\frac{V_0}{4}$:$e\left(\frac{V_0}{4}\right) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$From equation $(1)$,we have $eV_0 = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0})$. Substituting this into equation $(2)$:$\frac{hc}{4}(\frac{1}{\lambda} - \frac{1}{\lambda_0}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$Dividing both sides by $hc$:$\frac{1}{4\lambda} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{\lambda_0}$Rearranging the terms to solve for $\lambda_0$:$\frac{1}{\lambda_0} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{4\lambda}$$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$Therefore,$\lambda_0 = 3\lambda$.

$A$ 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 -

Similar Questions

Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of the cathode is $4 \ eV$. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely make the photocurrent zero?

Photoelectrons are emitted with maximum velocity $v$ when light of frequency $3f$ is incident on a photosensitive material of work function $2hf$. If the frequency of the incident light is $4.25f$,the maximum velocity of the emitted photoelectrons is ($h$ = Planck's constant).

In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

The work functions of Silver and Sodium are $4.6 \ eV$ and $2.3 \ eV$,respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is :

The work function of tungsten is $4.50 \, eV$. The wavelength of the fastest electron emitted when light whose photon energy is $5.50 \, eV$ falls on the tungsten surface is ......... $\mathring{A}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module