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Home » Monochromatic radiation of wavelength from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be The source is replaced by an iron source and its line irradiates the same photocell. Predict the new stopping voltage.
Monochromatic radiation of wavelength 640.2 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 \mathrm{~V} . The source is replaced by an iron source and its 427.2 \mathrm{~nm} line irradiates the same photocell. Predict the new stopping voltage.
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
Monochromatic radiation of wavelength 640.2 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 \mathrm{~V} . The source is replaced by an iron source and its 427.2 \mathrm{~nm} line irradiates the same photocell. Predict the new stopping voltage.

Wavelength of the monochromatic radiation is given as \lambda=640.2 \mathrm{~nm}=640.2 \times 10^{-9} \mathrm{~m}

Stopping potential of the neon lamp is also given as V_{0}=0.54 \mathrm{~V}

Charge on an electron, e=1.6 \times 10^{-19} \mathrm{C}

Planck’s constant, h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}

The following relation can be derived from photoelectric effect,
eV-{o}=hv-\phi_{o}

Work function of the metal is,
\Phi_{0}=\mathrm{hv}-\mathrm{eV}_{0}

=(\mathrm{hc} / \lambda)-\mathrm{eV}_{0}

=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{640.2 \times 10^{-9}}-1.6 \times 10^{-19} \times 0.54

=3.093 \times 10^{-19}-0.864 \times 10^{-19}

=2.229 \times 10^{-19} \mathrm{~J}

The radiation emitted by the iron source has a wavelength of \lambda^{\prime}=427.2 \mathrm{~nm}=427.2 \times 10^{-9} \mathrm{~m}

Let \mathrm{V}_{0} ‘ be the new stopping potential

Therefore, eV_{0^{'}}=(hc/\lambda^{'})-\phi_{o}

    \[=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{427.2 \times 10^{-9}}-2.229 \times 10^{-19}\]

=4.63 \times 10^{-19}-2.229 \times 10^{-19}

=2.401 \times 10^{-19} \mathrm{~J}

\mathrm{V}_{0}^{6}=2.401 \times 10^{-19} \mathrm{~J} / 1.6 \times 10^{-19} \mathrm{~J}

=1.5 \mathrm{eV}

The new stopping potential =1.50 \mathrm{eV}

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