A) 4125 nm
B) 412.5 nm
C) 41.250 nm
D) 4 nm
Correct Answer: B
Solution :
Energy of photon, \[E=\frac{hc}{\lambda }\] If energy E is expressed in eV and wavelength (in \[\overset{\text{o}}{\mathop{\text{A}}}\,\]), then \[E=\frac{12375}{\lambda \text{(}\overset{\text{o}}{\mathop{\text{A}}}\,\text{)}}eV\] \[\therefore \] \[\lambda =\frac{12375}{E}\overset{\text{o}}{\mathop{\text{A}}}\,=\frac{12375}{3}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=4125\overset{\text{o}}{\mathop{\text{A}}}\,=412.5\,\,nm\]
You need to login to perform this action.
You will be redirected in
3 sec
