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

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)]`

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To solve the problem, we need to find the de Broglie wavelength of a proton fired towards a nucleus with charge \( Q = 120e \) and a closest approach of \( r = 10 \, \text{fm} \). We will use the principles of energy conservation and the de Broglie wavelength formula. ### Step-by-Step Solution: 1. **Identify Given Values:** - Charge of the nucleus \( Q_1 = 120e \) - Charge of the proton \( Q_2 = e \) - Closest approach \( r = 10 \, \text{fm} = 10 \times 10^{-15} \, \text{m} \) ...
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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) .

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) .

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