Solved Examples: Structure of Atom | Mock Tests for JEE Main and Advanced 2025 PDF Download

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 ×10⁻³⁴ Js and c = 3.0 ×10⁸ ms⁻¹)   (IIT JEE-2013)
(a) 2.816 ×10⁻⁷ m
(b) 6.500 ×10⁻⁷m
(c) 8.500 ×10⁻⁷m
(d) 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 ×10⁻³⁴ Js
c = 3.0 ×10⁸ ms⁻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)
(a) h²/4π²m a₀²
(b) h²/16π²m a₀²
(c) h²/32π²m a₀²
(d) 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.