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A proton is first from very loward a nucleus with charge {Q}={

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A proton is first from very loward a nucleus with charge $Q = 120 e,$ where e is the nucles The de Brogle wavelength (in unit of fin) of the proton at its start is (tke the proton mass , $m_{p} = (5 / 3) \times 10^{- 27} k g h / s = 4.2 \times 10^{- 15} J s / C,$
$\frac{1}{4 π s _{0}} = 9 \times 10^{9} m / F, 1 f m = 10^{- 15} m$

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A proton is first from very loward a nucleus with charge

Q = 120 e,

where e is the nucles The de Brogle wavelength (in unit of fin) of the proton at its start is (tke the proton mass ,

m_{p} = (5 / 3) \times 10^{- 27} k g h / s = 4.2 \times 10^{- 15} J s / C,

\frac{1}{4 π s _{0}} = 9 \times 10^{9} m / F, 1 f m = 10^{- 15} m

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Text solutionVerified

Solution:

1

cos in K.E. of proton = gain in potential energy
of the proton - nucleus system

\frac{1}{2} m ν^{2} = \frac{1}{4 π s _{0}} \frac{q _{1} q _{2}}{r}

∴ \frac{p ^{2}}{2 m} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r}

:. (1)/(2m)((h^(2))/(h^(2))) = (1)/(4 pi epsilon_(0))(q_(1) q_(2))/(r )

:. lambda = sqrt((4 pi epsilon_(0) r , h^(2))/(q_(1) q_(2) (2m))) = 7 fm

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Question Text	A proton is first from very loward a nucleus with charge $Q = 120 e,$ where e is the nucles The de Brogle wavelength (in unit of fin) of the proton at its start is (tke the proton mass , $m_{p} = (5 / 3) \times 10^{- 27} k g h / s = 4.2 \times 10^{- 15} J s / C,$ $\frac{1}{4 π s _{0}} = 9 \times 10^{9} m / F, 1 f m = 10^{- 15} m$
Answer Type	Text solution:1
Upvotes	150