Question

Calculate the energy of an electron in the second bohr orbit of an excited hydrogen atom

Answer 1

Answer:  $5.35\times 10^{-19}J$

Explanation:

$E=\frac{hc}{\lambda}$

E= energy

$\lambda$ = Wavelength of radiation

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\times \frac{1}{n^2}\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n$ = energy level = 2

Z = atomic number (for hydrogen Z=1)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=1.09\times 10^7m^{-1}\frac{1}{2^2}$

$\lambda=3.7\times 10^{-7}m$

$E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{3.7\times 10^{-7}m}$

$E=5.35\times 10^{-19}J$

Thus the energy of an electron in the second bohr orbit of an excited hydrogen atom is $5.35\times 10^{-19}J$

Answer 2

Answer:

En = - Rh [ 1/n^2]

= -2. 18 × 10^-18 × (1/2^2)

= -2. 18 × 10^-18 × (1/4)

= -5. 44 × 10^-19 J

Explanation:

JUST REFER THE ANSWER.