JEE Mains · Chemistry · STD 11 - 2. structure of atom
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is \(....\,\times 10^{-19} \,J\) [Given: The threshold frequency of platinum is \(1.3\) \(\times 10^{15} \,s ^{-1}\) and \(h =6.6 \times 10^{-34} \,J \,s\).]
- A \(32.1\)
- B \(0.624\)
- C \(8.58\)
- D \(976\)
Answer & Solution
Correct Answer
(C) \(8.58\)
Step-by-step Solution
Detailed explanation
\(W = h v\) \(=6.6 \times 10^{-34} \times 1.3 \times 10^{15}\) \(=8.58 \times 10^{-19} \,J\)
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