As per the law of conservation of energy, the energy associated with an incident photon (E) must be equal to the sum of its kinetic energy and the work function (W_{0}) of the radiation.

E = W_{0} + K.E

⇒ W_{0} = E – K.E

Energy of incident photon

\( (E)= \dfrac{hc}{\lambda }\)

Where,

c denotes the velocity of the radiation

h is Planck’s constant

λ is the wavelength of the radiation

Substituting these values in the expression for E:

\(E=\dfrac{(6.626\times 10^{-34}Js)(3\times 10^{8}ms^{-1})}{(256.7\times 10^{-9}m)}\)

\( =7.744\times 10^{-19}\, J\\ \\ \)

\( =\dfrac{7.744\times 10^{-19}}{1.602\times 10^{-19}}eV\\ \\ \)

\( E=4.83\, eV\)

The potential that is applied to the silver is transformed into the kinetic energy (K.E) of the photoelectron.

Hence,

K.E = 0.35 V

K.E = 0.35 eV

Therefore, Work function, W_{0} = E – K.E

= 4.83 eV – 0.35 eV

= 4.48 eV

Answered by Pragya Singh | 1 year agoAnswer the following questions:-

**(a)** How many sub-shells are associated with n = 4?

**(b)** How many electrons will be present in the sub-shells having m_{s} value of \(- \dfrac{1}{2}\) for n = 4?

Indicate the number of unpaired electrons in:

**(a)** P

**(b) **Si

**(c)** Cr

**(d)** Fe

**(e)** Kr

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

**(i) **2s and 3s,

**(ii)** 4d and 4f,

**(iii)** 3d and 3p

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?