Text Solution Solution : When a charged particle of charge q, mass m is accelerated under a potential difference V, Let v be the velocity acquired by particle. Then <br> `qV=1/2mv^2 or mv=sqrt(2qVm)` <br> (i) `lambda=h/(mv)=h/(sqrt(2qVm)) or lambda prop 1/(sqrt(qm))` <br> `:. (lambda_e)/(lambda_p)=sqrt((q_pm_p)/(q_em_e))=sqrt((exx1837m_e)/(exxm_e))gt1` <br> So, `lambda_egtlambda_p`, i.e., greater value of de-broglie wavelength is associated with electron as compared to proton. <br> (ii) Momentum of particle, `p=mv=sqrt(2qVm)` <br> `:. p prop sqrt(qm)`, <br> Hence `(p_e)/(p_p)=sqrt((q_em_e)/(q_pm_p))=sqrt(e/exx(m_e)/(1837m_e)) lt 1` <br> So, `p_e lt p_p`, i.e., lesser momentum is associated with electron as compared to proton.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. a.) Find the energy and momentum of each photon in the light beam. c.) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? Given, m = 4.74 × 1014Hz (a) Energy of a photon, E = hv = 6.63 × 10-34× 4.74 × 1014J= 3.14 × 10-19J. Momentum of each photon, p (momentum) = hλ = 6.63 × 10-34632.8 × 10-9 = 1.05 × 10-27 kg ms-1 (b) Power emitted, P = 9.42 mW = 9.42 x 10–3 W n = PE = 9.42 × 10-3W3.14 × 10-19J = 3 × 1016 photons/sec. (c) Velocity of hydrogen atom = Momentum 'p' of H2 atom (mv)Mass of H2 atom(m) v = 1.05 × 10-271.673 × 10-27ms-1 = 0.63 ms-1. Thus, the hydrogen atom travel at a speed of 0.63 m/s to have the same momentum as that of the photon. |