A) the ratio of Planck's constant of electronic charge
B) work function
C) Planck's constant
D) charge of electron
Correct Answer: D
Solution :
\[{{I}_{0}}\] or. \[\frac{{{I}_{0}}}{2}.\] or, \[{{60}^{\circ }},\] On comparing the above equation with the straight line equation, i.e y = \[I=\left( \frac{{{I}_{0}}}{2} \right){{\left( \cos 60 \right)}^{2}}\] The slope of \[\text{=}\frac{{{\text{I}}_{\text{0}}}}{\text{2}}\frac{\text{1}}{\text{4}}\text{=}\frac{{{\text{I}}_{\text{0}}}}{\text{8}}\text{=0}\text{.125}{{\text{I}}_{\text{0}}}\] vs v is \[I\alpha \frac{1}{{{\lambda }^{4}}}\] Similarly \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{\lambda }_{2}}}{{{\lambda }_{1}}} \right)}^{4}}\] or, \[\Rightarrow \] Thus, slope of \[\frac{A}{{{I}_{2}}}={{\left( \frac{330}{880} \right)}^{4}}={{\left( \frac{3}{8} \right)}^{4}}=\frac{81}{4096}\] vs v is \[{{I}_{2}}=\frac{4096}{81}A=\left( 50.557 \right)A\] \[{{I}_{A}}\approx {{10}^{-10}}m\] \[{{I}_{N}}\approx {{10}^{-15}}m\]
You need to login to perform this action.
You will be redirected in
3 sec
