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Question: Light of wavelength \( 500nm \) is incident on a metal with a work function \( 2.28eV \) . The de Br...

Light of wavelength 500nm500nm is incident on a metal with a work function 2.28eV2.28eV . The de Broglie wavelength of the emitted electron is:
(A) ⩽2.8×10−9m\leqslant 2.8 \times {10^{ - 9}}m
(B) <2.8×10−10m< 2.8 \times {10^{ - 10}}m
(C) <2.8×10−9m< 2.8 \times {10^{ - 9}}m
(D) ⩾2.8×10−9m\geqslant 2.8 \times {10^{ - 9}}m

Explanation

Solution

Hint : Use Einstein’s photoelectric equation to find the de Broglie’s wavelength of the emitted electron. Einstein’s photoelectric equation is given by, E=w0+Kmax⁡E = {w_0} + {K_{\max }} where, EE is the incident energy of t light w0{w_0} is the work function of the metal and Kmax⁡{K_{\max }} is the kinetic energy of the electron after leaving the metal surface.

Complete Step By Step Answer:
We know that Einstein's photoelectric equation says that the incident energy of the light on a metal is used to free the electron in the metal and the rest energizes the electron in the form of kinetic energy. The minimum energy to free an electron from a metal surface is known as the work function of that metal. Now we know Einstein's photoelectric equation is given by, E=w0+Kmax⁡E = {w_0} + {K_{\max }} where, is the work function of the metal and is the kinetic energy of the electron after leaving the metal surface.
Now, here we have given, the wavelength of the incident electron is λ=500nm\lambda = 500nm and the work function of the metal w0=2.28eV{w_0} = 2.28eV . Hence, putting the values we get,
hcλ=w0+Kmax⁡\dfrac{{hc}}{\lambda } = {w_0} + {K_{\max }} [Since, we know E=hcλE = \dfrac{{hc}}{\lambda } ]
Or, 6.6×10−34×3×108500×10−9×1.6×10−19=2.28+Kmax⁡\dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{500 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} = 2.28 + {K_{\max }} [Where, Planck’s constant h=6.625×10−34h = 6.625 \times {10^{ - 34}} , c=3×108c = 3 \times {10^8} ]
So, Kmax⁡=0.195eV{K_{\max }} = 0.195eV
Now, we know that an electron leaving the surface of the conductor cannot have kinetic energy more than that. Now, we know that the momentum of a particle is given by, p=2mKp = \sqrt {2mK}
So, here we have the particle as electron and kinetic energy of the electron is, Kmax⁡=0.195eV{K_{\max }} = 0.195eV mass of the electron is, m=9.31×10−31kgm = 9.31 \times {10^{ - 31}}kg . putting the values we get,
p=2×9.31×10−31×0.195×1.6×10−19p = \sqrt {2 \times 9.31 \times {{10}^{ - 31}} \times 0.195 \times 1.6 \times {{10}^{ - 19}}} [ 1eV=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J ]
Or, p=2.4102×10−25p = 2.4102 \times {10^{ - 25}}
Now, we know de Broglie’s equation is given as, λ=hp\lambda = \dfrac{h}{p} .
Hence, putting the values we get, λ=6.625×10−342.4102×10−25\lambda = \dfrac{{6.625 \times {{10}^{ - 34}}}}{{2.4102 \times {{10}^{ - 25}}}}
Or, λ=2.8×10−9\lambda = 2.8 \times {10^{ - 9}}
Hence, this the minimum value of wavelength that an electron can have while leaving the surface of the metal. Since, kinetic energy is inversely proportional to the wavelength of the particle.
Hence, the value of wavelength will be, λ⩾2.8×10−9m\lambda \geqslant 2.8 \times {10^{ - 9}}m
Hence, option ( D) is correct.

Note :
An electron leaving the surface of the conductor cannot have kinetic energy more than E−w0E - {w_0} . Now, we know that the momentum of a particle is given by, p=2mKp = \sqrt {2mK} . if K⩽Kmax⁡K \leqslant {K_{\max }} then we can say, p⩽pmax⁡p \leqslant {p_{\max }} . Now, p=hλp = \dfrac{h}{\lambda } so we can say, hλ⩽pmax⁡\dfrac{h}{\lambda } \leqslant {p_{\max }} or, λ⩾hpmax⁡\lambda \geqslant \dfrac{h}{{{p_{\max }}}} . Hence, if kinetic energy has the maximum value, then for that maximum kinetic energy wavelength will be minimum.