Question

State photo electric effect. The work function for caesium atom is 1.9 ev.Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. (c) If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer 1

Chemistry/structure of atom

AnSwer :-

Photoelectric effect :-

The phenomenon of ejection of electrons from the surface of metal when light of suitable frequency strikes it is called a photoelectric effect. The ejected electrons are called photoelectrons. It is given that the work function (W0) for caesium atom is 1.9eV.

(a) from the expression,

$\sf W_0 = \dfrac{hc}{ \lambda _{0} }$

we get,

$\sf \lambda_0 = \dfrac{hc}{W_0 }$

where, $\lambda_0$ = threshold wavelength, h = Planck's constant, c = velocity of radiation.

substituting the values in the given expression of ($\lambda_0$):

$\tt \lambda_0 = \dfrac{(6.626 \times {10}^{ - 34} Js)(3.0 \times {10}^{8} {ms}^{ - 1}) }{1.9 \times 1.602 \times {10}^{ - 19} J }$

$\tt\lambda_0 = 6.53 \times {10}^{ - 7} m$

Hence, the threshold wavelength is 653nm.

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(b) From the expression

$\sf W_0 = hv_0$

we get,

$\tt v_0 = \dfrac{W_0}{h}$

where, v$_0$ = Threshold frequency, h = Planck's constant

substituting the values in the given expression of v$_0$

$\tt v_0 = \dfrac{1.9 \times 1.602 \times {10}^{ - 19}J }{6.626 \times {10}^{ - 34} Js}$

$\tt(1eV = 1.602 \times {10}^{ - 19} J)$

$\tt v = 4.593 \times {10}^{14} \: {s}^{ - 1}$

Hence, the threshold frequency of radiation (v$_0$) is 4.593 × 10¹⁴ s.

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(c) According to the question

wavelength used in irradiation ($\lambda$) = 500nm

Kinetic energy = h(v - v$_0$)

$\tt = hc \bigg( \dfrac{1}{ \lambda} - \dfrac{1}{ \lambda_0} \bigg)$

$\tt = (6.626 \times {10}^{ - 34} Js)(3.0 \times {10}^{8} {ms}^{ - 1} ) \bigg( \dfrac{ \lambda_0 - \lambda}{ \lambda \lambda_0 } \bigg)$

$\tt = (1.9878 \times {10}^{ - 26} Jm) \bigg[ \dfrac{(653 - 500) {10}^{ - 9}m }{(653)(500) {10}^{ - 18} {m}^{2} } \bigg]$

$\tt= \dfrac{(1.9878 \times {10}^{ - 26})(153 \times {10}^{9}) }{(653)(500)} J$

$\tt = 9.3149 \times {10}^{ - 20} J$

Kinetic energy of the ejected photoelectron = $\sf = 9.3149 \times {10}^{ - 20} J$

since,

$\tt K.E = \dfrac{1}{2} m {v}^{2} = 9.3149 \times {10}^{ - 20} J$

$\tt v = \sqrt{ \dfrac{2(9.3149 \times {10}^{ - 20} J)}{9.10939 \times {10}^{ - 31} Kg} }$

$\tt = \sqrt{2.0451 \times {10}^{11} {m}^{2} {s}^{ - 2} }$

v = 4.52 × 10⁵ m/s

Hence, the velocity of the ejected photoelectron (v) is 4.52 × 10⁵ m/s.

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