Find the accelerating potential `(V)` that must be impurated to a belium atom so that its wavelegth is `5 Å (1 amu = 1.67 xx 10^(-24)g)`

Answer

A, B, D

Explanation

`lambda = (h)/(sqrt(2Em)) = `…….(i) `[{:(lambda=(h)/(p)=(h)/(mv)","E=(1)/(2)mv^(2)),(mv^(2)=2E),(mxxmv^(2)=2Exxm),(mv=sqrt(2Em)),(p=sqrt(2Em)):}]`
`m = (He) = 4 xx 1.67 xx 10^(-27) kg`
Therefore subsituting value in equation (i)
`5 xx 10^(-10) m = ((6.62 xx 10^(-34))^(2) J s)/(sqrt(2). sqrt(E) xx (4 xx 1.67 xx 10^(-27))^(1//2)`
`:. E = ((6.62 xx 10^(-34))^(2))/(2 xx 4 xx 1.67 xx 10^(-27) xx (5 xx 10^(-10))^(2))`
`=(45.8244 xx 10^(-68))/(334 xx 10^(-27)) = 0.1312 xx 10^(-21)`
`E = e V`
`= (E)/(e ) = (0.082 xx 10^(-2)) V `
`= 8.2 xx 10^(-4) V`