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

**(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 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?