Question

The ground state energy of hydrogen atom is 13.6 electron volt if an electron makes a transition from an energy level -1.5 one electron volt to -3.4 electron volt then calculate the wavelength of the spectral line emitted and name of the series of hydrogen spectrum

Answer 1

Explanation:

E= -3.4 for n=2

E= -1.5 for n=3

so substitute formula and get the answer. see the image attatched

Answer 2

The wavelength of the emitted light is 6.6 × 10^-7 m and the wavelength lies in the balmer series .

Energy in an orbital = -13.6 × Z²/n²

Z = 1 , for hydrogen atom

n - the orbital number

E(n1) = energy of the electron at n1 orbital = -3.4 eV.

n1 = √(13.4/3.6) = 2

E(n2) = energy of the electron at n2 orbital = -1.5 eV.

n2 = √(13.4/1.5) = 3

λ - wavelength of the transition from n1 to n2

Using formula , 1/λ = RZ²(1/(n1)² - 1/(n2)²) to find the transition wavelength

Z = 1 , n1 = 2 , n2 = 3

=> 1/λ = R (1/4 - 1/9)

=> 1/λ = R (5/36)

=> λ = 36/5R

λ = 6.6 × 10^-7 m

The wavelength lies in the balmer series.