A proton is fired from very for 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, hle = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)]`

Explanation

`r = closest distance = 10 fm`
From energy conservation, we have
`K_i+ U_i = K_f +U_f`
or K+ 0 = 0+ (1)/(4pi_0). (q_1q_2)/(r ) or K= (1)/(4piepsilon_0. ((120 e) (e )/(r )
de- Broglie wavelength,
`lambda = (h)/(sqrt2Km)
Substituting the given values in above two equations, we get `lambda = 7xx10^(-15) m 7 fm
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