Dual Nature of Matter

JEE Main level, single correct, medium

126 of 200

Question 126 of 200

The threshold frequency of a metal with work function 6.63eV6.63eV is :

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Answer: (D) 1.6×1015Hz1.6\times {10}^{15}Hz

The formula to calculate the work function of the metal is given by
ϕ0=hv0...(1){\phi }_{0}=h{v}_{0}...\left(1\right)
From equation (1), it follows that
6.63×1.6×10−19=6.63×10−34v0⇒v0=1.6×10−1910−34⇒v0=1.6×1015Hz6.63\times 1.6\times {10}^{-19}=6.63\times {10}^{-34}{v}_{0} \Rightarrow {v}_{0}=\dfrac{1.6\times {10}^{-19}}{{10}^{-34}} \Rightarrow {v}_{0}=1.6\times {10}^{15}Hz

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