Question

find out the wavelength of the electron orbiting in the first excited state in hydrogen atom​

Answer 1

Answer;

Radius of ground state of hydrogen atom=0.53A*=0.53*10^-10m

according to de Broglie relation 2*pi*r=n/\

For ground state n=1

2*3.14*0.53*10^-10=1*/\

therefore, /\=3.32*10^-10m

=3.32A*

Answer 2

$\huge{\text{\underline{Question:-}}}$

Find out the wavelength of the electron orbiting in the first excited state in hydrogen atom.

$\huge{\text{\underline{Solution:-}}}$

★ According to De Broglie's equation

$\implies$λ = h / (mv)

★ Where:-

• h is the Planck's constant and it's value is 6.626 X 10⁻³⁴ J
• m is the mass of the electron and it's value is 9.1 X 10⁻³¹ kg.

The velocity of an electron in the first orbit of the hydrogen atom.

• v ’ = 2.18 x 10⁶ m/s

The velocity of an electron in the second orbit (n = 2) of the hydrogen atom.

$\implies$v = v ’/ n

$\implies$v = 2.18 x 10⁶ / 2

$\implies$v = 1.09 x 10⁶ m/s

$\implies$mv = 9.919 x 10⁻²⁵

Therefore, mv = 9.919 x 10⁻²⁵

★ According to the equation

λ = (6.626 × 10⁻³⁴) / (9. 919 × 10⁻²⁵) m

λ = 6.680 x 10⁻¹⁰ m

Hence, the wavelength λ = 6.680 Å

__________________________________________