Calculate the energy of H-atom orbital (J atom-1 ) whose angular momentum is 4.2178 × 10-34 kg m 2 / sec.

Question

Calculate the energy of H-atom orbital (J atom-1 ) whose angular momentum is 4.2178 × 10-34 kg m 2 / sec. (A) -5.45 × 10-19 (B) -1.36 × 10-19 (C) -87.8 × 10-19 (D) -34.88 × 10-18

Tardigrade Admin · Accepted Answer

m v r=(n h/2 π)=4.2178 × 10-34 n=(4.2178 × 10-34 × 2 × 3.14/6.6 × 10-34)=4 Thus, E4=(-2.18 × 10-18/42) J atom-1=-1.36 × 10-19 J atom-1

Q. Calculate the energy of H-atom orbital ( $J a t o m^{- 1}$ ) whose angular momentum is $4.2178 \times 1 0^{- 34} k g m^{2} / sec$ .

$- 5.45 \times 1 0^{- 19}$

$- 1.36 \times 1 0^{- 19}$

$- 87.8 \times 1 0^{- 19}$

$- 34.88 \times 1 0^{- 18}$

$m v r = \frac{nh}{2 π} = 4.2178 \times 1 0^{- 34}$

$n = \frac{4.2178 \times 1 0 ^{- 34} \times 2 \times 3.14}{6.6 \times 1 0 ^{- 34}} = 4$

Thus, $E_{4} = \frac{- 2.18 \times 1 0 ^{- 18}}{4 ^{2}} J a t o m^{- 1} = - 1.36 \times 1 0^{- 19} J a t o m^{- 1}$

Q. Calculate the energy of H-atom orbital (Jatom−1 ) whose angular momentum is 4.2178×10−34kgm2/sec.

−5.45×10−19

−1.36×10−19

−87.8×10−19

−34.88×10−18