Question

Light of wavelength 4000 Å falls on the surface of cesium. Calculate the energy of the photoelectron emitted. The critical wavelength for photoelectric effect in cesium is 6600 Å. ​

Answer 1

Aиѕωєr —

Energy of the photoelectron emitted = $\bold{1.95\times 10^{-19}J}$

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Giνєи —

Wavelength of light = 4000A°

The critical wavelength for photoelectric effect for cesium = 6600A°

Tσ Fiиd —

The energy of the photoelectron emitted .

Sσℓυтiσи –

Maximum energy of photoelectron emitted = Initial energy - photoelectric energy

$\qquad\qquad\red{\bigstar}\underline{\boxed{\large\bf{ E_{max}= E_{i}-E_{p}}}}\red{\bigstar}$

$\qquad\qquad\rm{\implies E_{max}=\dfrac{hc}{\lambda_{1}}-\dfrac{hc}{\lambda_{2}}}$

✪ Hєrє —

$\boxed{\begin{array}{cc} \\ \qquad \sf {\leadsto h = 6.6 \times 10^{ - 34} joule - second} \\ \\ \sf {\leadsto c = 3 \times {10}^{ 8} m {s}^{ - 1}} \qquad \: \: \: \: \: \: \: \\ \\ \sf{\leadsto \lambda_{1}=4000A°} \qquad \qquad \: \: \: \: \: \\ \: \: \: \sf{=4000\times 10^{-10}m} \\ \\ \sf{\leadsto \lambda_{2}=6600A°} \qquad \qquad \: \: \: \: \: \\ \: \: \: \sf{=6600\times 10^{-10}m} \\ \end{array}}$

✪ Oи Pυттiиg Vαℓυєѕ —

$\qquad\qquad\rm{\implies E_{max}=6.6\times 10^{-34}\times 3\times 10^{8}\left(\dfrac{1}{4000\times 10^{-10}}-\dfrac{1}{6600\times 10^{-10}}\right)}$

$\qquad\qquad\rm{\implies E_{max}=19.878\times 10^{-26}\left(\dfrac{2600\times 10^{-10}}{264\times 10^{-15}}\right)}$

$\qquad\qquad\rm{\implies E_{max}=195.768\times 10^{-21}}$

$\qquad\qquad\rm{\implies E_{max}= 1.95\times 10^{-19}J}$

$\large{\bull}$ Hence , $\bf{1.95\times 10^{-19}J}$ Energy of photoelectron emitted .

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