A) 950 nm
B) 820 nm
C) 580 nm
D) 440 nm
Correct Answer: D
Solution :
Given, \[E=2.8\,\,eV=2.8\times 1.6\times {{10}^{-19}}\] Energy \[E=\frac{hc}{\lambda }\] \[\therefore \] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.8\times 1.6\times {{10}^{-19}}}\] \[=440\times {{10}^{-9}}m=440\,nm\]
You need to login to perform this action.
You will be redirected in
3 sec
