TS EAMCET · Chemistry · Structure of Atom
In a photoelectric effect experiment, the kinetic energy of an emitted electron is \(1.986 \times 10^{-19} \mathrm{~J}\), when a radiation of frequency \(1.0 \times 10^{15} \mathrm{~s}^{-1}\) hits the metal. What is the threshold frequency of the metal \(\left(\right.\) in s \(\left.^{-1}\right)\) ? (Planck's constant \(=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}\) )
- A \(7.0 \times 10^{14}\)
- B \(5.8886 \times 10^{14}\)
- C \(7.0 \times 10^{-15}\)
- D \(7.0 \times 10^{15}\)
Answer & Solution
Correct Answer
(A) \(7.0 \times 10^{14}\)
Step-by-step Solution
Detailed explanation
Given, kinetic energy of an emitted electron \[ =1.986 \times 10^{-19} \mathrm{~J} \] Frequency of radiation \(=1 \times 10^{15} \mathrm{~s}^{-1}\) Threshold frequency, \(v_0=\) ? From equation,…
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