Question

Calculate the energy of H-atom orbital ($J\, atom^{-1}$ ) whose angular momentum is $4.2178 \times 10^{-34} kg\, m ^{2} / sec$.

• A. $-5.45 \times 10^{-19}$
• B. $-1.36 \times 10^{-19}$
• C. $-87.8 \times 10^{-19}$
• D. $-34.88 \times 10^{-18}$

Answer 1

Answer: (B)

$m v r=\frac{n h}{2 \pi}=4.2178 \times 10^{-34}$

$n=\frac{4.2178 \times 10^{-34} \times 2 \times 3.14}{6.6 \times 10^{-34}}=4$

Thus, $E_{4}=\frac{-2.18 \times 10^{-18}}{4^{2}} J\, atom^{-1}=-1.36 \times 10^{-19} \,J\, atom^{-1}$