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A proton is fired from very far away to...

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)`.

Text Solution

Verified by Experts

`((9xx10^9)(120e)(e))/(10xx10^(-15))=(p^2)/(2m)`
`lamda=(h)/(p)`
`lamda^2=(h^2)/(p^2)=(h^2xx10xx10^(-15))/(2mxx9xx10^(-9)xx120e^2)`
`(4.2xx4.2xx10^(-30)xx10xx10^(-15))/(2((5)/(3))xx10^(-27)xx9xx10^(-9)xx120)`
`(42xx42)/(36)xx10^(-30)`
`implieslamda=7xx10^(-15)m=7fm`
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