Solved Examples on Atomic Structure

Question: 1) Energy of an electron is given by E = -2.178×10^-18 J Z²/n². Wavelength of light required to excite an electron in an hydrogen atom from level n =1 to n = 2 will be (h = 6.62 ×10⁻³⁴Js and c = 3.0 ×10⁸ms⁻¹)   (IIT JEE-2013)

1) 2.816 ×10⁻⁷ m

2) 6.500 ×10⁻⁷m

3) 8.500 ×10⁻⁷m

4) 1.214 ×10⁻⁷m

Answer: d

Solution:

We know that energy of an electron can be calculated using Planck relation

E = hc/λ     …………… (i)

But in the question it is given that

E = -2.178×10^-18 J Z² (1/n₁² -1/n₂²) …………(ii)

On comparing both the equations we get,

hc /λ =-2.178×10^-18 J Z²(1/n₁² -1/n₂²) ………….(iii)

On substituting the values

h =6.62 ×10⁻³⁴Js

c = 3.0 ×10⁸ms⁻¹

n₁= 1 & n₂= 2

In equation (iii) and calculating for the value of λ we get,

λ =1.214 ×10⁻⁷m

Hence, the correct option is d.

Question: 2) The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of He gas at –73 ^oC is “M” times that of the de Broglie wavelength of Ne at 727^oC. M is  (IIT JEE-2013) 

Solution:

 

Question: 3) The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a⁰ is Bohr radius] (IIT JEE 2012)

1) h²/4π²m a₀²

2) h²/16π²m a₀²

3) h²/32π²m a₀²

4) h²/64π²m a₀²

Answer: c

Solution:

According to Bohr’s postulate,

Electrons can exist only in those orbital around the nucleus, for which the angular momentum is integral multiple of  h/2π i.e.

mvr  = n h/2π

or,

 v = n h/2πmr

So, Kinetic Energy of electrons = ½ m (n h/2πmr)²

Radius of the orbital i.e. r_n = (a₀×n²)/Z

For, n=2 & z =1

r = 4 a₀

So,

Kinetic Energy of electrons = ½ m [4 h²/(4π²m²(4 a₀))]² = h²/32π²m a₀²

Hence, the correct option is C.