A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of `10fm` to the nucleus. The de - Broglie wavelength (in units of fm) of the proton at its start is take the proton mass, `m_p = 5//3xx10^(-27) kg, h//e = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)`.

Options

A. 7
B. 9
C. 11
D. 13

Answer

A

Explanation

(a) : As `(1)/(4piepsilon_(0))((120e)(e))/(10xx10^(-15))=(P^(2))/(2m_(p))` . . .(i)
where p is the momentum of the proton and `m_(p)` is the mass of the proton
de Broglie wavelength of proton,
`lamda=(h)/(p)orp=(h)/(lamda)`
Substituting this value of p in equation (i), we get
`(1)/(4piepsilon_(0))(120e^(2))/(10xx10^(-15))=(h^(2))/(2lamda^(2)m_(p))`
`lamda^(2)=(4piepsilon_(0)xx10xx10^(-15)xxh^(2))/(2m_(p)xx120e^(2))`
Substituting the given numerical values, we get
`lamda^(2)=(1xx10xx10xx^(-15)xx4.2xx10^(-15)xx4.2xx10^(-15))/(9xx10^(9)xx2xx(5)/(3)xx10^(-27)xx120)`
`(42xx42xx10^(-30)xx10^(-14))/(36xx10^(-14))=49xx10^(-30)`
`lamda=7xx10^(-15)m=7fm`