AP EAMCET · Chemistry · Structure of Atom
A light of frequency \(1.6 \times 10^{16} \mathrm{~Hz}\) when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of \(1.0 \times 10^{16} \mathrm{~Hz}\) falls on the same plate. The threshold frequency \(\left(v_0\right)\) of the metal in \(\mathrm{Hz}\) is
- A \(1 \times 10^{15}\)
- B \(4 \times 10^{15}\)
- C \(3 \times 10^{15}\)
- D \(4 \times 10^{13}\)
Answer & Solution
Correct Answer
(B) \(4 \times 10^{15}\)
Step-by-step Solution
Detailed explanation
When \(1.6 \times 10^{16} \mathrm{~Hz}\) frequency falls on metal plate than double kinetic energy obtained. \[ h\left(1.6 \times 10^{16}-v_0\right)=2 \mathrm{~K} . \mathrm{E} \longrightarrow \text { Eq. (i) } \] When \(1.0 \times 10^{16} \mathrm{~Hz}\) frequency falls on same…
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