If velocity of a particle is 3 times of that of electron and ratio of de-broglie wavelength of particle to that of electron is `1.814 xx 10^(–4)` . The particle will be -

Options

A. neutron
B. deuteron
C. alpha particle
D. tritium

Answer

1

Explanation

`lambda_(p) = 1.814 xx 10^(-4) xx lambda_(e )`
`(h)/(m_(p)v_(p)) = 1.814 xx 10^(-4) xx (h)/(m_(e) v_(e))`
`rArr m_(p) = (v_(e ))/(v_(p)) xx (1)/(1.814 xx 10^(4)) xx m_(e )`
`rArr m_(p) = (1)/(3) xx (10^(-4))/(1.814) xx 9.1 xx 10^(-31) = 1.67 xx 10^(-27)`