A) \[4125\,\overset{o}{\mathop{A}}\,\]
B) \[3000\,\overset{o}{\mathop{A}}\,\]
C) \[6000\,\overset{o}{\mathop{A}}\,\]
D) \[2062.5\,\overset{o}{\mathop{A}}\,\]
Correct Answer: B
Solution :
| The maximum wavelength above which no photoelectron can emit from metal surface is called cut-off wavelength and is given by work function \[=\frac{hc}{\text{cut-off wavelength}}\] |
| or cut-off wavelength \[=\frac{hc}{\text{work function}}\] |
| \[\therefore \] \[{{\lambda }_{0}}=\frac{hc}{W}\] ...(i) |
| Given, \[h=6.6\times {{10}^{-34}}\,J-s\] |
| \[c=3\times {{10}^{8}}\,m/s\] |
| \[W=4.125\,eV=4.125\times 1.6\times {{10}^{-19}}\,J\] |
| Substituting the given values in Eq. (i), we get |
| \[{{\lambda }_{0}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4.125\times 1.6\times {{10}^{-19}}}\overset{\text{o}}{\mathop{\text{A}}}\,\] |
| \[=3\times {{10}^{-7}}\,m=3000\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\] |
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