Dual Nature Of Light Question 323
Question: If the energy of a photon corresponding to a wavelength of 6000 Å is $ 3.32\times {{10}^{-19}}J $ , the photon energy for a wavelength of 4000 Å will be [DPMT 2004]
Options:
A) 1.4 eV
B) 4.9 eV
C) 3.1 eV
D) 1.6 eV
Show Answer
Answer:
Correct Answer: C
Solution:
$ E=\frac{hc,}{\lambda },\Rightarrow ,\frac{E_{1}}{E_{2}}=\frac{{\lambda_{1}}}{{\lambda_{2}}},\Rightarrow ,\frac{3.32\times {{10}^{-19}}}{E_{2}}=\frac{4000}{6000} $
$ \Rightarrow E_{2}=4.98\times {{10}^{-19}}J=3.1,eV. $
BETA
