TS EAMCET · Chemistry · Structure of Atom
The kinetic energy of electrons emitted, when radiation of frequency \(1.0 \times 10^{15} \mathrm{~Hz}\) hits a metal, is \(2 \times 10^{-19} \mathrm{~J}\). What is the threshold frequency of the metal (in Hz)? \(\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)\)
- A \(3.5 \times 10^{15}\)
- B \(3.3 \times 10^{14}\)
- C \(6.97 \times 10^{15}\)
- D \(6.97 \times 10^{14}\)
Answer & Solution
Correct Answer
(D) \(6.97 \times 10^{14}\)
Step-by-step Solution
Detailed explanation
Given : - Kinetic energy \(=2 \times 10^{-19} \mathrm{~J}\) Radiation of frequency \(=1.0 \times 10^{15} \mathrm{~Hz}\) \(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\) According to Einstein's equation Kinetic energy…
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