Sankalp JEE Full Test-1 Question-38 Solution

Question 38. The threshold frequency of a metal with work function 6.63 eV is :

(1) 16 \times 10^{15} ~Hz

(2) 16 \times 10^{12} ~Hz

(3) 1.6 \times 10^{12} ~Hz

(4) 1.6 \times 10^{15} ~Hz

Answer (4)

Explanation:

\phi_{0}=h v_{0} 6.63 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_{0} v_{0}=\frac{1.6 \times 10^{-19}}{10^{-34}} v_{0}=1.6 \times 10^{15} ~Hz
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