A) 1
B) 3
C) 4
D) 6
Correct Answer: C
Solution :
Specific charge of electron \[\frac{e}{m}\,=1.8\times {{10}^{11}}\,Ck{{g}^{-1}}\] Maximum kinetic energy of photoelectrons \[\frac{1}{2}m{{v}^{2}}_{\max }\,=e{{V}_{s}}\] Where \[{{V}_{s}}\]is the stopping potential \[{{V}_{s}}\,=\frac{mv_{\max }^{2}}{2e}=\frac{v_{\max }^{2}}{2(e/m)}\] \[=\frac{{{(1.2\,\times {{10}^{6}})}^{2}}}{2\times 1.8\times {{10}^{11}}}\,=0.4\,\times 10\,=4\,V\]You need to login to perform this action.
You will be redirected in
3 sec