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  4. The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

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Q. The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491 nm$ is $0.710 V$. When the incident wavelength is changed to a new value, the stopping potential is $1.43 V$. The new wavelength is :

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Solution:

$\frac{ hc }{\lambda}=\phi+ eV _{ s }$
$\frac{1240}{491}=\phi+0.71 \ldots$(1)
$\frac{1240}{\lambda}=\phi+1.43 \ldots $(2)
$\therefore \lambda=382 \,nm$