InterviewSolution
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InterviewSolution
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The rediation emitted when an electron jumps from `n = 3 to n = 2` orbit in a hydrogen atom falls on a metal to produce photoelectron. The electron 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 (a) the kinetic energy of the electrons, (b) Work function of the metal , and (c) wavelength of radiation. |
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Answer» When charged particle moves perpendicular to a magnetic field , then magneticfield provides necessary centripetel force for circle path of radius`r` given by `(m v^(2))/( r) = q v B` `implies m v = q B r` As momentum, `p = mv = sqrt(2 m E_(K))`, where `E_(K)` is kinetic energy , therefore `sqrt(2 m E_(K)) = qBr` `:. E_(k) = ((qBr)^(2))/(2 m)` `{1.6 xx 10^(-19) xx ((1)/(320)) xx 10^(-3)}^(2)/(2 xx 9.1 xx 10^(-31) J` `= 1.374 xx 10^(-19)J = (1.374 xx 10^(-19))/(1.6 xx 10^(-19)) e V` `= 0.86 e V` Energy of photon released due to transition from `n = 3 to n = 2` in hydrogen atom. `epsilon = Delta E` `= Rhc ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (13.6 e V) ((1)/(2^(2)) - (1)/(3^(2)))` `= 1.89 e V` Work function of metal `W = epsilon - E_(k) = 1.89 - 0.86 = 1.03 e V` c. Wavelength of emitted radition (photon) is given by `Delta E = (hc)/(lambda)` `implies lambda = (hc)/(Delta E) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(1.89 xx 1.6 xx 10^(-19))` `= 6.567 xx 10^(-7) m = 6567Å` |
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