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If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximatelyA. `0.3A`B. `3A`C. `30A`D. `10A` |
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Answer» Correct Answer - A We have `KE=(p^2)/(2m_e)=(hc)/(lamda_(min))` `P=sqrt((2hcm_e)/(lamda_(min)))` For `lamda_(min)=10A` `lamda_(de Broglie)cong0.3A` |
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