When a photon of energy 4.25 , eV strikes the | vedclass

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

Question

(B) From Einstein's photoelectric equation: $4.25 = \phi_{A} + T_{A}$  --- $(1)$$4.70 = \phi_{B} + T_{B}$  --- $(2)$Given: $T_{B} = T_{A} - 1.50 \, eV

Vedclass · Accepted Answer

The work function of $B$ is $4.20 \, eV$

Vedclass · Answer

(B) From Einstein's photoelectric equation: $4.25 = \phi_{A} + T_{A}$  --- $(1)$$4.70 = \phi_{B} + T_{B}$  --- $(2)$Given: $T_{B} = T_{A} - 1.50 \, eV$De-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.Thus,$\frac{\lambda_{A}}{\lambda_{B}} = \sqrt{\frac{T_{B}}{T_{A}}}$.Given $\lambda_{B} = 2\lambda_{A}$,so $\frac{1}{2} = \sqrt{\frac{T_{B}}{T_{A}}} \Rightarrow \frac{1}{4} = \frac{T_{B}}{T_{A}} \Rightarrow T_{A} = 4T_{B}$.Substituting $T_{A} = 4T_{B}$ into the given relation: $T_{B} = 4T_{B} - 1.50 \Rightarrow 3T_{B} = 1.50 \Rightarrow T_{B} = 0.50 \, eV$.Then $T_{A} = 4 	imes 0.50 = 2.0 \, eV$.From $(1)$,$\phi_{A} = 4.25 - 2.0 = 2.25 \, eV$.From $(2)$,$\phi_{B} = 4.70 - 0.50 = 4.20 \, eV$.Therefore,the work function of $B$ is $4.20 \, eV$.

Similar Questions

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

At an incident radiation frequency of $v_1$,which is greater than the threshold frequency,the stopping potential for a certain metal is $V_1$. At frequency $2 v_1$,the stopping potential is $3 V_1$. If the stopping potential at frequency $4 v_1$ is $n V_1$,then $n$ is

If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on a photosensitive surface. If only $1 \%$ of the incident photons emit photoelectrons,calculate the number of photoelectrons emitted per second from a surface area of $1 \ cm^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module