Solved Numerical Problem No.3 ( Unit-19: Atomic Spectra); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

An electron drops from the second energy level to the first energy level within an excited hydrogen atom (a) determine the energy of the photon emitted (b) calculate the frequency of the photon emitted (c) calculate the wavelength of the photon emitted. (Answer: (a) 10.2 eV (b) 2.5 ｘ 10¹⁵ Hz (c) 120 nm)

Given Data:

Atom: Hydrogen

Excited state = n = 2

Lower state = p = 1

To Find:

(a) Energy of the photon emitted = E = ?

(b) Frequency of the photon emitted = f = ?

Solution:

(a) Energy of the photon emitted = E = ?

The energy of the photon emitted is given by:

E =

$\E_n$ -

$\E_p$ -------------(1)

Where

$\E_n$ is the energy of an electron in an excited state and

$\E_p$ is the energy of an electron in the lowest state.

$\E_n$ = -

$\frac {E_0}{n^2}$

Where

$\E_0$ = constant = 13.6 eV and n = 2 So,

$\E_n$ = -

$\frac {13.6 eV}{2^2}$

$\E_n$ = -

$\frac {13.6 eV}{4}$

$\E_n$ = - 3.4 eV

and

$\E_p$ = -

$\frac {E_0}{p^2}$

Where

$\E_0$ = constant = 13.6 eV and p = 1 So,

$\E_p$ = -

$\frac {13.6 eV}{1^2}$

$\E_p$ = -

$\frac {13.6 eV}{1}$

$\E_p$ = - 13.6 eV

Now putting values in Equation (1)

E =

$\E_n$ -

$\E_p$ = - 3.4 eV - (- 13.6 eV ) = 10.2 eV

E = 10.2 eV -------------------Ans.1

(b) Frequency of the photon emitted = f = ?

We know that

E = hf

f =

$\frac {E}{h}$

By putting values of E = 10.2 eV = 10.2 ｘ 1.6 ｘ 10⁻¹⁹ J and Plank's Constant = h = 6.62 ｘ 10⁻³⁴ m² kg s⁻¹

f =

$\frac {10.2 ｘ 1.6 ｘ 10⁻¹⁹ J}{6.62 ｘ 10⁻³⁴ m² kg s⁻¹}$

By simplifying we get

f = 2.46 ｘ 10¹⁵ Hz -------------------Ans.2

we know that v = f λ, where v = c = 3 ｘ 10⁸ m s⁻¹

λ =

$\frac {3 ｘ 10⁸ m s⁻¹}{2.46 ｘ 10¹⁵ Hz }$

λ = 1.219 ｘ 10⁻⁷ m

λ = 121.9 ｘ 10⁻⁹ m

λ = 121.9 nm ------------------Ans.3

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