The radiation corresponding to 3→2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10⁻⁵ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to 0.8eV, 1.8eV, 1.1eV, 1.6eV

mv=qBR
For the largest circular path, the momentum of the electron is maximum, which implies that the kinetic energy of the electron is also maximum.
The kinetic energy of the electron is given by:
KE = 1/2 mv²
The radius of the circular path is given by:
r = mv/qB
mv = qBr
KE = (qBr)²/2m
The energy of the photon is given by:
E = hc/λ
where h is Planck's constant, c is the speed of light, and λ is the wavelength of the radiation.
For the 3→2 transition in hydrogen atom, the wavelength of the radiation is given by:
1/λ = R(1/2² - 1/3²) = R(5/36)
where R is Rydberg's constant.
λ = 36/5R
E = hcR(5/36) = (5hcR)/36
The energy of the photon is given by:
E = KE + W
where W is the work function of the metal.
W = E - KE
W = (5hcR)/36 - (qBr)²/2m
Substituting the values:
h = 6.63 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
R = 1.097 × 10⁷ m⁻¹
q = 1.6 × 10⁻¹⁹ C
B = 3 × 10⁻⁵ T
r = 10 × 10⁻³ m
m = 9.1 × 10⁻³¹ kg
W = (5 × 6.63 × 10⁻³⁴ × 3 × 10⁸ × 1.097 × 10⁷)/36 - ((1.6 × 10⁻¹⁹ × 3 × 10⁻⁵ × 10 × 10⁻³)²)/(2 × 9.1 × 10⁻³¹)
W = 3.03 × 10⁻¹⁹ J - 1.06 × 10⁻¹⁹ J
W = 1.97 × 10⁻¹⁹ J
W = 1.97 × 10⁻¹⁹ J / (1.6 × 10⁻¹⁹ J/eV)
W ≈ 1.23 eV
The closest option is 1.1 eV.