In a photoelectric experiment,the potential d | vedclass

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

Question

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by:$K_{max} = hf - \phi$whe

Vedclass · Accepted Answer

$h(f_1 - f_0)$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by:$K_{max} = hf - \phi$where $h$ is Planck's constant,$f$ is the frequency of incident light,and $\phi$ is the work function of the metal.In the photoelectric effect,the stopping potential $V$ is related to the maximum kinetic energy by the relation $K_{max} = eV$,where $e$ is the charge of an electron.Thus,$eV = hf - \phi$,or $V = \frac{h}{e}f - \frac{\phi}{e}$.From the graph,when $f = f_0$ (threshold frequency),the stopping potential $V = 0$. Therefore,$0 = \frac{h}{e}f_0 - \frac{\phi}{e}$,which implies $\phi = hf_0$.Substituting this into the expression for $K_{max}$ at frequency $f_1$:$K_{max} = hf_1 - hf_0 = h(f_1 - f_0)$.Therefore,the correct option is $C$.

Similar Questions

Which of the following statements is correct in the case of the photoelectric effect?

The photoelectric threshold wavelength for potassium (work function being $2 \ eV$) is ........... $nm$.

In a photocell,the incident wavelength is $\lambda$. The maximum speed of the emitted photoelectrons is $u$. If the incident wavelength is changed to $3\lambda / 4$,then the maximum speed of the emitted photoelectrons will be:

Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $R$ by a perpendicular magnetic field having a magnitude $B$. The work function of the metal is (where,symbols have their usual meanings)

Vedclass Test Series

Exam Paper Generator

Online Exam Module