The correct answer is (A): 1.36 eV
Explanation:
\(\frac{hc}{λ}\) - Ø = ΚΕ........(i)
R =\(\frac{ mv}{Bq}\) = \(\sqrt{\frac{2m(KE)}{Bq }}\) ........(ii)
Putting the values,
Ø ≃ 1.36 eV
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