A) 5525 V
B) 320 V
C) 6200 V
D) 3250 V
Correct Answer: C
Solution :
Potential difference\[V=\frac{hc}{e\lambda }\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-10}}}\] = 6200 voltYou need to login to perform this action.
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