JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Modern Physics

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 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass,

m_{p} = (5/3) \times 1 0^{- 27} kg

;

h / e = 4.2 \times 1 0^{- 15} J . s / C

;

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} m / F

;

1 fm = 1 0^{- 15} m

)

Answer: 7

Solution

Energy conservation:

\frac{1}{2} m_{p} v^{2} = \frac{k Q e}{r _{min}} = \frac{9 \times 1 0 ^{9} \times 120 e \times e}{10 \times 1 0 ^{- 15}}

. Solving for

v

v^{2} = \frac{2 \times 9 \times 1 0 ^{9} \times 120 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 5/3 ) \times 1 0 ^{- 27} \times 10 \times 1 0 ^{- 15}}

. The de Broglie wavelength is

λ = \frac{h}{m _{p} v} = \frac{( h / e ) \times e}{m _{p} v} = \frac{4.2 \times 1 0 ^{- 15} \times 1.6 \times 1 0 ^{- 19}}{m _{p} v}

. After calculation,

λ \approx 7 \times 1 0^{- 15}

m = 7 fm.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Modern Physics

Solution

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