Question

If the total energy of an electron in hydrogen like atom in excited state is –3.4 eV, then the de Broglie wavelength of the electron is

• A. 6.6 × 10^–10
• B. 3 × 10^–10
• C. 5 × 10⁹
• D. 9.3 × 10^–12

Answer 1

Answer: (A)

6.6 × 10^–10

Answer 2

Total energy (E_n) = $\frac{E_{1}}{n^{2}}$ ⇒ –3.4 eV = $\frac{- 13.6}{n^{2}}$

⇒ n² = $\frac{- 13.6}{- 3.4}$ = 4 ∴ n = 2

The velocity of electron in 2^nd orbit = $\frac{V_{1}}{2}$ =

$\frac{2.18 \times 10^{6}}{2}$ m/sec,

λ = $\frac{h}{mv}$ = $\frac{6.625 \times 10^{- 34} \times 2}{9.1089 \times 10^{- 31} \times 2.18 \times 10^{6}}$ = 6.6 × 10^–12 m

= 6.6 × 10^–10 cm