(i) Velocity |
(ii) Wavelength |
(iii) Frequency |
(iv) Energy |
A) (ii) only
B) (ii) and (iv)
C) (ii), (iii) and (iv)
D) (iv) only
Correct Answer: C
Solution :
[c] e/m waves shown in figure A has higher wavelength in comparison to e/m waves shown in figure B. Thus these waves also differ in frequency and energy, \[v=\frac{v}{\lambda }\] \[\Rightarrow {{E}_{1}}=\frac{hc}{{{\lambda }_{1}}}\] \[\Rightarrow {{E}_{2}}=\frac{hc}{{{\lambda }_{2}}}{{\lambda }_{1}}>{{\lambda }_{2}}\Rightarrow {{E}_{1}}<{{E}_{2}}\]You need to login to perform this action.
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