Structure Of Atom Ques 64
64. If the energy of a photon is given as $3.03 \times 10^{-19} $ $J$ then, the wavelength $(\lambda)$ of the photon is :
[2000]
(a) $6.56 $ $nm$
(b) $65.6 $ $nm $
(c) $656 $ $nm$
(d) $0.656 $ $nm$
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Answer:
Correct Answer: 64.(c)
Solution:
- (c) The energy of photon,
$E=\frac{h c}{\lambda}=3.03 \times 10^{-19}$
or $\lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3.03 \times 10^{-19}}$
or $\lambda=\frac{19.878}{3.03} \times 10^{-7}=6.56 \times 10^{-7} m=656$ $ nm$
BETA
