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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.

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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 727oC. 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. rn = (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.

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