**Question 3 : Calculate the wavelength in angstroms of electromagnetic radiation emitted by a hydrogen atom which undergoes a transition between energy levels of -1.36**×10^{−19}**J and ‘-5.45 **×10^{−19}**J. (Take Planck??s constant h = 6.6 **×10^{′}^{−34}**Js)**

E1– E2 = hf = hc/λ

E1 = ** -1.36**×10^{−19}**J**

**E2 = 5.45 **×10^{−19}**J**

**6.6**×10^{′}^{−34}**Js**

**λ = ?**

Solution

E1– E2 = hf = hc/λ

E don’t have frequency so we going to solve for frequency first

Frequency can be calculated using the following equation

E2 – E1 = hf

F = __E2 – E1__

h

**-5.45 **×10^{−19}**J + 1.36**×10^{−19}**J = 6.6 **×10^{′}^{−34}**JsF making F the subject of the formula we have **

**F = -5.45 **

__×10__

^{−19}

__J + 1.36____×10__

^{−19}

__J__**= -619696969696969.69696969696969697 approximately is**

** 6.6 **×10^{′}^{−34}**Js**

= 6.2 X 10^{15}Hz

F = 6.2 X 10^{15}Hz

Now we want to look for wavelength

Which is E1– E2 = hc/λ

Λ = hc/ E1– E2

So we

E1 = -1.8eV

E2 = -4.0eV

-1.8 + 4.0 = 2.2ev

H = **6.6 **×10^{′}^{−34}**Js**

C = 3 .0 ×10^{′}^{8}**ms-1**

**6.6 **×10^{′}^{−34}**Js x **3 .0 ×10^{′}^{8}**ms-1 = 0.000000000000000000000000198**

Λ = —————————————————————————————————– =

2.2 x 3 .0 ×10^{′}^{8}**ms-1 = 0.000000000000000352**

**=0.0000000005625 approximately = **5.625 x 3 .0 ×10^{′}^{-10}**m answer **

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