`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will beA. `0.59 nm`B. `0.59 Å`C. `0.59 p m`D. `0.59 p m`

`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._

1.	`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will beA. `0.59 nm`B. `0.59 Å`C. `0.59 p m`D. `0.59 p m`
Answer» Correct Answer - D Suppose closet distance is `r`, according to conservation of energy. `400 xx 10^(3) xx 1.6 xx 10^(-19) = 9 xx 10^(9) ((ze)(2e))/(r )` `rArr 6.4 xx 10^(-14)` `= (9 xx 10^(9) xx (82 xx 1.6 xx 10^(-19))xx (2 xx 1.6 xx 10^(-19)))/(r )` `rArr r = 5.9 xx 10^(-13) m = 0.59p m`

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`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will beA. `0.59 nm`B. `0.59 Å`C. `0.59 p m`D. `0.59 p m`

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