1 Crore+ students have signed up on EduRev. Have you? 
Q.1. Let T1 and T2 be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to Bohr’s model of an atom, the ratio T1 : T2 is
(1) 4 : 1
(2) 4 : 9
(3) 9 : 4
(4) 1 : 4 (2022)
Answer (3)
En = E_{o }/ n^{2}
For first excited state: n = 2
For second excited state n = 3
_{}
Q.2. For which one of the following, Bohr model is not valid? (2020)
(a) Deuteron atom
(b) Singly ionised neon atom (Ne^{+})
(c) Hydrogen atom
(d) Singly ionised helium atom (He^{+})
Ans: b
Bohr Model is valid only for those atoms which have one electron in orbit.
Bohr Model
Q.2. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively: (2019)
(a) –3.4 eV, –3.4 eV
(b) –3.4 eV, –6.8 eV
(c) 3.4 eV, –6.8 eV
(d) 3.4 eV, 3.4 eV
Ans: c
In Bohr's model of H atom
∴ K.E. = 3.4 eV
U = –6.8 eV
Q.3. αparticle consists of : (2019)
(a) 2 protons and 2 neutrons only
(b) 2 electrons, 2 protons and 2 neutrons
(c) 2 electrons and 4 protons only
(d) 2 protons only
Ans: a
αparticle is nucleus of Helium which has two protons and two neutrons.
Q.4. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom,is : (2018)
(a) 1 : 1
(b) 1 : –1
(c) 2 : –1
(d) 1 : –2
Ans: b
Q.5. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is(e + De). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then De is of the order of [Given mass of hydrogen mh = 1.67 × 10^{–27} kg] (2017)
(a) 10^{–23} C
(b) 10^{–37} C
(c) 10^{–47} C
(d) 10^{–20} C
Ans: b
We know, a hydrogen atom has one electron and one proton
so, net charge on each hydrogen atom = (e + ∆e  e) = ∆e
so, electrostatic force ,
gravitational force,
where M_{h} denotes mass of hydrogen
a/c to question,
gravitational force = electrostatic force
we know, K = 9 × 10^{9} Nm^{²}/C^{²}, G = 6.67 × 10^{11} Nm^{²}/Kg^{²} , M_{h} = 1.67 × 10^{27} Kg
so, 9 × 10^{9} × ∆e^{²} = 6.67 × 10^{11} × (1.67 × 10^{27})^{²}
or, ∆e^{²} = 6.67 × 10^{11} × 1.67 × 1.67 × 10^{54}/9 × 10^{9}
or, ∆e = 1.43767 × 10^{37 }C
Q.6. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is : (2017)
(a) 1
(b) 4
(c) 0.5
(d) 2
Ans: b
For last line of Balmer : n_{1} = 2 & n_{2} = ∞
For last line Lyman series : n_{1} = 1 & n_{2} = ∞
Q.7. When an αparticle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze' its distance of closet approach from the nucleus depends on m as : (2016)
(a) m
(b) 1/m
(c) 1/√m
(d) 1/m^{2}
Ans: b
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
This is the required distance of closest approach to alpha particle from the nucleus.
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
This is the required distance of closest approach to alpha particle from the nucleus.
Q.8. Given the value of Rydberg constant is 10^{7} m^{1}, the wave number of the last line of the Balmer series in hydrogen spectrum will be : (2016)
(a) 2.5 x 10^{7}m^{1}
(b) 0.025 x 10^{4}m^{1}
(c) 0.5 x 10^{7}m^{1}
(d) 0.25 x 10^{7}m^{1}
Ans: d
Rydberg constant, r = 10^{7} m^{1}
For last line in Balmer series,
n_{2} = ∞; n_{1} = 2
We know,
Q.9. Consider 3^{rd} orbit of He^{+} (Helium), using nonrelativistic approach, the speed of electron in this orbit will be [given K = 9 x 10^{9} constant, Z = 2 and h(Planck's Constant) = 6.6 x 10^{34} J s] (2015)
(a) 3.0 x 10^{8} m/s
(b) 2.92 x 10^{6} m/s
(c) 1.46 x 10^{6} m/s
(d) 0.73 x 10^{6} m/s
Ans: c
Q.10. If radius of thenucleus is taken to be R_{Al}, then the radius ofnucleus is nearly : (2015)
(a)
(b)
(c)
(d)
Ans: c
Radius of the nucleus goes as R ∝ A^{1/3}, where A is the atomic mass. If R_{Te} is the radius of the nucleus of telurium atom and R_{Al} is the radius of the nucleus of aluminium atom we have
There is no Question for NEET 2021
157 videos452 docs213 tests
