A) 1.5V
B) 3.85V
C) 2.35V
D) 15.4V
Correct Answer: A
Solution :
\[{{\lambda }_{1}}=\frac{12375}{{{E}_{1}}(eV)}\overset{o}{\mathop{\text{A}}}\,=1000\overset{o}{\mathop{\text{A}}}\,\] \[\therefore \] \[{{E}_{1}}=12.375\,eV\] ...(i) Similarly, \[{{E}_{2}}=\frac{12375}{{{\lambda }_{2}}(\overset{o}{\mathop{\text{A}}}\,)}\,eV\] \[=\frac{12375}{2000}\,=6.1875\,eV\] ?.(ii) Now, \[{{E}_{1}}-{{W}_{0}}=e{{V}_{s}}\] ...(iii) and \[{{E}_{2}}-{{W}_{0}}=e{{V}_{s}}\] ...(iv) Hence, \[12.375-{{W}_{0}}=7.7\,eV\] ...(v) and \[6.1875-{{W}_{0}}=e{{V}_{s}}\] ...(vi) Solving, we get \[{{V}_{s}}=1.5\,V\]You need to login to perform this action.
You will be redirected in
3 sec