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.
b.) How many photons per second, on the average arrive at a target irradiated by this beam?c.) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Given,
Wavelength of monochromatic light, λ = 632.8 nm = 632.8 x 10–9
m
∴ Frequency, v = cλ = 3 × 108632.8 × 10-9Hz
= 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
Now, P = nE
n = PE = 9.42 × 10-3W3.14 × 10-19J = 3 × 1016 photons/sec.
Thus, these many number of protons arrive at the target.(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.