Matter waves (DC Broglie waves)

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Question

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

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Solution

$E = W + {K_{\max }};$
$\frac{{1240}}{{500}} = 2.28 + {K_{\max }}\$
2.48 = 2.28 + K_max ;
K_max = 0.2 eV ; V = 0.2 V
$\lambda \geqslant \frac{{12.27}}{{\sqrt V }}; \geqslant \frac{{12.27}}{{\sqrt {0.2} }}\,\mathop A\limits^0 \geqslant 2.8 \times {10^{ - 9}}m\$
Note : Velocity of ejected electrons may be less than V_max so λ is greater than 2.8 × 10^–9 m

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Matter waves (DC Broglie waves)

Remember concepts with our Masterclasses.

Light of wavelength 500 nm is incident on a met­al with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

2.48 = 2.28 + Kmax ;Kmax = 0.2 eV ; V = 0.2 VNote : Velocity of ejected electrons may be less than Vmax so λ is greater than 2.8 × 10–9 m

Ready to Test Your Skills?

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Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is