Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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 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.
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 727oC. M is (IIT JEE-2013)
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
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.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
