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The energines `E_(1) and E_(2)` of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., `lamda_(1) and lamda_(2)` will beA. `lamda_(1) = (1)/(2) lamda_(2)`B. `lamda_(1) = lamda_(2)`C. `lamda_(1) = 2 lamda_(2)`D. `lamda_(1) = 4 lamda_(2)` |
Answer» Correct Answer - C `E = hv = h (c)/(lamda) :. (E_(1))/(E_(2)) = (lamda_(2))/(lamda_(1))` i.e., `(lamda_(1))/(lamda_(2)) = (E_(2))/(E_(1)) = (50eV)/(25 eV) = 2 or lamda_(1) = 2 lamda_(2)` |
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