In 1924, de-Broglie proposed that every particle possesses wave properties with a wavelength , `lambda` given by `lambda = h/(mv)` where m is the mass of the particle, `v` is its velocity and h is Planck's constant. The de-Broglie prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction , a phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy of an electron accelerating by a potential V as :
`1/2 mv^2 = eV "where" 1eV = 1.6 xx 10^(-19) J, h = 6.6 xx 10^(-34) Js`.
The mass of a photon moving with velocity of light having wavelength same as that of an `alpha` - particle (mass = `6.6 xx 10^(-27)kg)` moving with velocity of `2.5 xx 10^2 ms^(-1)` is

Options

A. `7.92 xx 10^(-21) kg`
B. `5.5 xx 10^(-33) kg`
C. `5.65 xx 10^(-31) kg`
D. `7.92 xx 10^(-28) kg`

Answer

B

Explanation

Wavelength of `alpha`-particles
`lambda(alpha"- particle") = h/(mv) = (h)/(6.6 xx 10^(-27) xx 2.5 xx 10^2)`
`lambda("photon") = h/(m xx 9.0 xx 10^8)`
Since `lambda(alpha- "particle") = lambda` (photon)
`h/(6.6 xx 10^(-27) xx 2.5 xx 10^2) = h/(m xx 3.0 xx 10^8)`
`m = (6.6 xx 10^(-27) xx 2.5 xx 10^2)/(3.0 xx 10^8)`
`=5.5 xx 10^(-33) kg`.