Question

please give correct option.be fast guys ..im requsting u​

Answer 1

Answer:

is the 2.2*10^-11m

HERE IS THE ANSWER

Answer 2

The energy The energy of the electron in hydrogen-like atom in nth orbit is

$E_n=\frac{Z^2Rhc}{n^2}$

Rhc = 1 rydberg

The ionisation energy

E∞ - E₁ = Z²Rhc = 4 rydberg

∴ Z² = 4rydberg / Rhc = 4rydberg / 1rydberg = 4

Hence, Z = 2.

The energy required to excite the electron from n=1 to n=2 is given by

E₂ - E₁ = -Z²Rhc / 2² - ( -Z²Rhc/1² )

         = Z²Rhc( 1 - 1/4 )

         = 3/4 Z²Rhc = 3/4 × 4rydberg

         = 3 rydberg

If λ is the wavelength of radiation emitted , then

hc / λ = 3 Rydberg , i.e, λ = hc / 3 rydberg

Hence, λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (3 × 2.2 × 10⁻¹⁸)

λ = 301.4 × 10⁻¹⁰ m

Radius = (ε₀h² / πme²) / Z

           = (Radius of first Bohr orbit of hydrogen ≥ n) / Z  

           = (5 × 10⁻¹¹) / 2

          = 2.5 × 10⁻¹¹ m

Option (C) is correct.