Photoelectric emission is observed from a surface for frequency ${v_1}$ and ${v_2}$ of the incident radiation $\left( {{{\rm{v}}_1} > {{\rm{v}}_2}} \right).$ If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1∶k, then the threshold frequency ${{\rm{v}}_0}$ is given by:

$\frac{{{\rm{k}}{{\rm{v}}_1} - {{\rm{v}}_2}}}{{k - 1}}$

$\frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{{k - 1}}$

$\frac{{{\rm{k}}{{\rm{v}}_2} - {{\rm{v}}_1}}}{{k - 1}}$

$\frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{k}$

NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Hard

1 of 1

Application +4 | -1

Photoelectric emission is observed from a surface for frequency

{v_1}

and

{v_2}

of the incident radiation

\left( {{\mathrm{v}_1} > {\mathrm{v}_2}} \right).

If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1∶k, then the threshold frequency 

{\mathrm{v}_0}

 is given by:

View answer and explanation

Correct option: B

\frac{{\mathrm{k}{\mathrm{v}_1} - {\mathrm{v}_2}}}{{k - 1}}

Option B (

\frac{{\mathrm{k}{\mathrm{v}_1} - {\mathrm{v}_2}}}{{k - 1}}

) is correct because it represents the threshold frequency derived from the photoelectric effect equations and the given ratio of kinetic energies.