The radiation emitted, when an electron jumps from n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1 320 T in a radius of 10 - 3 m . Find the work function of the metal

Question

The radiation emitted, when an electron jumps from n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1 320 T in a radius of 10 - 3 m . Find the work function of the metal, 1 . 03 eV, 1 . 89 eV, 0 . 86 eV, 2 . 03 eV

MARKS App · Accepted Answer

E 3 - E 2 = 13.6 1 2 2 - 1 3 2 = 13.6 × 5 36 = 1.89 e V Photoelectron with K E m a x is moving on a circular path. r = m v q B m v = q B r P = q B r = 1.6 × 10 - 19 × 1 3200 × 10 - 3 = 1 2 × 10 - 24 = 5 × 10 - 25 k g m s - 1 The energy of photoelectron = K E m a x = p 2 2 m = 25 × 10 - 50 2 × 9.1 × 10 - 31 × 1.6 × 10 - 19 = 0.86 e V Now, using the Einstein equation hν = ϕ + K E m a x 1.89 = 0.86 + ϕ ⇒ ϕ = 1.03 e V

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