A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal is 2.13 eV. Calculate 

(i) the energy of the photon (eV), 

(ii) the kinetic energy of the emission, and 

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

Asked by Pragya Singh | 1 year ago |  110

1 Answer

Solution :-

(i)

Energy of the photon (E)= \(h\nu =\dfrac{hc}{\lambda }\)

Where, h denotes Planck’s constant, whose value is 

\(6.626\times 10^{-34}\,Js\)

c denotes the speed of light = \(3\times 10^{8}\,m/s\lambda\)

= wavelength of the photon =\(4\times 10^{-7}\,m/s\)

Substituting these values in the expression for E:

\(E=\dfrac{(6.626\times 10^{-34})(3\times 10^{8})}{4\times 10^{-7}}\)

\( =4.9695\times 10^{-19}\, J\)

Therefore, energy of the photon = \(4.97\times 10^{-19}\, J\)

(ii)

The kinetic energy of the emission\(E_{k}\)can be calculated as follows:

\(=h\nu -h\nu _{0}\\ \\ =(E-W)eV\\ \\ =(\dfrac{4.9695\times 10^{-19}}{1.6020\times 10^{-19}})eV-2.13\, eV\\ \\ =(3.1020-2.13)eV\\ \\ =0.9720\, eV\)

Therefore, the kinetic energy of the emission = 0.97 eV.

(iii)

The velocity of the photoelectron (v) can be determined using the following expression:

\(\dfrac{1}{2}mv^{2} =h\nu -h\nu _{0}\\ \\ \Rightarrow v =\sqrt{\dfrac{2(h\nu -h\nu _{0})}{m}}\)

Where\((h\nu -h\nu _{0})\) is the K.El of the emission (in Joules) and ‘m’ denotes the mass of the photoelectron.

Substituting thes values in the expression for v:

\(v=\sqrt{\dfrac{2\times (0.9720\times 1.6020\times 10^{-19})J}{9.10939\times 10^{-31}kg}}\\ \\ =\sqrt{0.3418\times 10^{12}m^{2}s^{2}}\\ \\ \Rightarrow v=5.84\times 10^{5}ms^{-1}\)

Therefore, the velocity of the ejected photoelectron is\(5.84\times 10^{5}ms^{-1}\)

Answered by Pragya Singh | 1 year ago

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