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What should be the momentum (in gram cm per second) of a particle if its De Broglie wavelength is `1Å` and the value of `h` is `6.6252xx10^(-27)` erg second?A. `6.6252xx10^(-19)`B. `6.6252xx10^(-21)`C. `6.6252xx10^(-24)`D. `6.6252xx10^(-27)` |
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Answer» Given that `lambda = 1 Å =1 xx10^(-8)cm," "h=6.6252xx10^(-27)` erg second `"or" " "p=(6.6252xx10^(-27))/(1xx10^(-8))=6.6252xx10^(-19) "gram" cm//sec`. |
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