Light coming from a discharge tube filled wit | vedclass

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

Question

(D) The work function of the cathode surface is $\phi = 4 \ eV$.In a hydrogen discharge tube,the maximum energy of a photon emitted during the transit

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(D) The work function of the cathode surface is $\phi = 4 \ eV$.In a hydrogen discharge tube,the maximum energy of a photon emitted during the transition of an electron from higher energy levels to the ground state $(n=1)$ is given by the Rydberg formula,where the maximum energy corresponds to the transition from $n = \infty$ to $n = 1$,which is $13.6 \ eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by $K_{max} = E_{photon} - \phi$.Substituting the maximum photon energy,$K_{max} = 13.6 \ eV - 4 \ eV = 9.6 \ eV$.To make the photocurrent zero,the stopping potential $(V_s)$ must be applied such that $e|V_s| = K_{max}$.Therefore,$|V_s| = 9.6 \ V$.Since the stopping potential must be negative with respect to the cathode to stop the electrons,the anode voltage must be at least $-9.6 \ V$. Among the given options,$-10 \ V$ is the only value that is more negative than $-9.6 \ V$,which will effectively stop all photoelectrons.

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?

Similar Questions

$A$ metal surface is illuminated by light of wavelength $400 \ nm$. The kinetic energy of the emitted photoelectrons is found to be $1.68 \ eV$. The work function of the metal is ......... $eV$. $(hc = 1240 \ eV \cdot nm)$

In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

The work functions of cesium $(Cs)$ and lithium $(Li)$ metals are $1.9 \ eV$ and $2.5 \ eV$,respectively. If we incident light of wavelength $550 \ nm$ on these two metal surfaces,then the photoelectric effect is possible for which case?

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be

The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module