A certain metallic surface is illuminated with monochromatic light of wavelength, \lambda . The stopping potential for photo-electric current for this light is 3V 0 . If the same surface is illuminated with light of wavelength 2 \lambda , the stopping potential is V 0 . The threshold wavelength for this surface for photo-electric effect is

Dual Nature of Radiation and Matter question

2015 · Q129

Dual Nature of Radiation and Matter question

2015 · Q129

View written solutionFree

Correct answer: D

Stopping potential, $\phi$ = E - K_max

3eV₀ + $\phi$ = ${{{hc} \over \lambda }}$ ......(1)

eV₀ + $\phi$ = ${{{hc} \over 2\lambda }}$ ......(2)

Solving equation (1) and (2), we get

$\phi$ = ${{{hc} \over {4\lambda }}}$

So, threshold wavelength,

$\lambda$ _th = ${{{hc} \over \phi }}$ = ${{{hc} \over {{{hc} \over {4\lambda }}}}}$ = 4 $\lambda$

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Dual Nature of Radiation and Matter question

2015 · Q129

Dual Nature of Radiation and Matter question

2015 · Q129

View written solutionFree

Correct answer: D

Stopping potential, $\phi$ = E - K_max

3eV₀ + $\phi$ = ${{{hc} \over \lambda }}$ ......(1)

eV₀ + $\phi$ = ${{{hc} \over 2\lambda }}$ ......(2)

Solving equation (1) and (2), we get

$\phi$ = ${{{hc} \over {4\lambda }}}$

So, threshold wavelength,

$\lambda$ _th = ${{{hc} \over \phi }}$ = ${{{hc} \over {{{hc} \over {4\lambda }}}}}$ = 4 $\lambda$