Question

Radiation of wavelength 4500 Å is

incident on a metal having work function

2.0 eV. Due to the presence of a magnetic

field B, the most energetic photoelectrons

emitted in a direction perpendicular to

the field move along a circular path of

radius 20 cm. What is the value of the

magnetic field B?​

Answer 1

Given :

• Wavelength $(\lambda)$ = 4500Å or ${\sf 45 \times 10^{-8}\: m}$
• Work function $(\phi)$ = 2eV
• Radius of Circular path (r) = 20cm or 0.2m

To Find :

• Magnetic Field (B)

Formula Used :

$\implies\underline{\boxed{\sf K = E - \phi}}$

K = Kinetic Energy

E = Energy of photon

$\phi$ = Work function

$\implies\underline{\boxed{\sf p = mv = \sqrt{2mK}}}$

p = momentum

m = mass of electron

$\implies\underline{\boxed{\sf r = \dfrac{mv}{qB}}}$

q = 1.6 × ${\sf 10^{-19}\:C}$

Solution :

Kinetic Energy of electron -

$\implies{\sf K = \dfrac{hc}{\lambda}-\phi }$

$\implies{\sf \dfrac{6.6\times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-8}} - 2 \times 1.6 \times 10^{-19}}$

$\implies{\sf \dfrac{19.8 \times 10^{-18}}{45}- 3.2 \times 10^{-19}}$

$\implies{\sf 0.44 \times 10^{-18}- 0.32 \times 10^{-18}}$

$\implies{\bf K = 12 \times 10^{-16}\: J }$

Moment of Electron (p) :

$\implies{\sf p = mv = \sqrt{2mK} }$

m = mass of Electron = 9.1 × ${\rm 10^{31}\:kg}$

$\implies{\sf \sqrt{2 \times 9 \times 10^{-31} \times 12 \times 10^{-16}} }$

$\implies{\sf \sqrt{21.8 \times 10^{-46}} }$

$\implies{\bf p = 4.6 \times 10^{-23}\: kg.m/s }$

The most energetic photoelectronsbemitted in a direction perpendicular to the field move along a circular path

$\implies{\sf r = \dfrac{mv}{qB}}$

$\implies{\sf 0.2= \dfrac{4.6\times 10^{-23}}{1.6 \times 10^{-19}\times B} }$

$\implies{\sf B = \dfrac{4.6 \times 10^{-23}}{1.6 \times 10^{-19}\times 0.2}}$

$\implies{\sf B = \dfrac{4.6 \times 10^{-23}}{3.2 \times 10^{-19}}}$

$\implies{\bf B = 1.4 \times 10^{-4}\: T}$

Answer :

Value of Magnetic Field is ${\bf 1.4 \times 10^{-4}\:T}$