If a proton is accelerated through a potential difference of 1000V, then its de-Broglie wavelength is (given, `m_(p) = 1.67 xx 10^(-27)kg, h = 6.63 xx 10^(-34)J-s`)

Options

A. `9.1xx 10^(-13)m`
B. `9.1 xx 10^(-13) m`
C. `1.09 xx 10^(-18)m`
D. `1.09 xx 10^(+16)m`

Answer

A

Explanation

`lamd_("proton") = (h)/(p) = (0.286)/(sqrtV) (Å)`
(This formula is for protons.)
`:. lamda = (0.286)/(sqrt1000) xx 10^(-10) m ~~ 9.1 xx 10^(-13)m`