Dual Nature Of Light Question 230
Question: If we express the energy of a photon in KeV and the wavelength in angstroms, then energy of a photon can be calculated from the relation [AMU (Engg.) 1999]
Options:
A) $ E=12.4,h\nu $
B) $ E=12.4,h/\lambda $
C) $ E=12.4,/\lambda $
D) $ E=h\nu $
Show Answer
Answer:
Correct Answer: C
Solution:
Energy of photon $ E=\frac{hc}{\lambda } $ (Joules) $ =\frac{hc}{e\lambda }(eV) $
$ \Rightarrow ,\underset{(eV)}{\mathop{E}},=\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{1.6\times {{10}^{-19}}\times \lambda ({\AA})}=\frac{12375}{\lambda ({\AA})} $
$ \Rightarrow E(keV)=\frac{12.37}{\lambda ({\AA})}\approx \frac{12.4}{\lambda } $
BETA
