A) 0.402 eV
B) 0.204 eV
C) 0.306 eV
D) \[0.\text{6}0\text{3 eV}\]
Correct Answer: D
Solution :
\[E=hv=\frac{hc}{\lambda }\] \[=\frac{6\cdot 63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4000\times {{10}^{-10}}}\] \[=4\times 966\times {{10}^{-19}}J\] \[=\frac{4\cdot 966\times {{10}^{-19}}}{1\cdot 6\times {{10}^{-19}}}eV=3\cdot 103\,\,eV\] Energy of emitted photon\[=(3.103-2.5)eV\] \[\mathbf{=0}\mathbf{.603}\,\,\mathbf{eV}\]
You need to login to perform this action.
You will be redirected in
3 sec
