MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 8
1
2
3
4
5
6
7
8
9
Q.1
) Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionization Potential of hydrogen is 13.6 v and the mass of hydrogen atom is 1.67 × 10⁻²⁷ kg. The quantum number of the two levels in the emission
0%
a) n=1 , n=3
0%
b) n=2 , n=4
0%
c) n=1 , n=4
0%
d) n=3 , n=4
Explanation
When radiation falls on surface, it energy is utilized to overcome bonding energy and remaing as kinetic energy It can be given by mathamatical equation as Incidant radiation energy (hν) = Work function (φ) + Kinetic energy Given Work function = 1.82 ev and Photoelectron energy = 0.73 ev Sunstituting above values in eqation we get Energy of incidant radiation hν = 1.82 +.73 =2.25 ev Energy of Hydrogen atom orbit is 13.6/n2 Clculate values of energy for different energy elevel and subtract Corresponding energy level for 2.55 eV is n=2 and n=4Answer: (b)
Q.2
) Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is 1.67 × 10⁻²⁷ kg. In this transition change in the angular momentum of electron is (where h isPlank constanst )
0%
a)h / 2π
0%
b) h / π
0%
c)2h / π
0%
d)3h / 2π
Explanation
change in Angular momentum=(4h / 2π) - ( 2h / 2π)=h / π Answer: (b)
Q.3
)Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is 1.67 × 10⁻²⁷ kg. The recoil speed of emitting atom caussing that is at lest before the transitionis of the order of
0%
a) 1 cm /s
0%
b) 102 m / s
0%
c)104 m / s
0%
d)1 m /s
Explanation
According to conservation of momentum, momentum of proton=moentum of recoil atom ∴ h / λ=m &mew μ=h / mλ=E / mcAnswer: (d)
Q.4
Suppose the charge of a proton and an electron differslightly. One of them is–e, the other is (e + Δe). Ifthe net of electrostatic force and gravitational forcebetween two hydrogen atoms placed at a distanced (much greater than atomic size) apart is zero,then Δe is of the order of [Given mass of hydrogen mh = 1.67 × 10⁻²⁷ kg]
0%
a) a) 10–20 C
0%
b) 10–23 C
0%
c) 10–37 C
0%
d) 10–47 C
Explanation
FE = FG Charge on each hydrogen atom = -e+e+Δe = Δe Answer:(c)
Q.5
The energy released by fission of one atom of 92U235 is 200 MeV. The number of fission required per second to produce a power of 1kW is ..[ AFMC 2001]
0%
a)3.125×109
0%
b)3.125×1012
0%
c)3.125×1013
0%
d)3.125×1011
Explanation
Energy released in joule per fission is=200× 106 ×1.6× 10⁻¹⁹Energy released in joule per fission is=320×10⁻¹³ J=320×10⁻¹⁶ kJIf 'n' is the number of fission per second then n=total energy per second / energy per fission n=1kW/ 320×10⁻¹⁶ n=3.125×1013 Answer:(c)
Q.6
Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. What would be the mass of U-235 consumed per day in the fission reactor of power 1MW approximately
0%
a) 10 kg
0%
b) 100kg
0%
c) 1 g
0%
d) 10-2g
Explanation
Amount of energy released per fission is 200 MeV=200(1.6×10⁻¹³=320-13 Require energy per sec=106 J Number of atoms required per second=required energy / energy released per fission Number of atoms required per second=106 / 320-13 Number of atoms required per second=1 / 32×10⁻¹⁸ Weight of fissile material required per second will be Answer: (c)
Q.7
) The kinetic energy of an electron which is accelerated through a potential difference of 100V is [ AFMC 2010]
0%
a)6.626 × 103 watt
0%
b) 1.16 × 10⁻³⁴ J
0%
c)418.6 cal
0%
d)1.6 × 10⁻¹⁷ J
Explanation
K.E.=q . v=100 × 1.6 ×: 10-19 J =1.6 × 10⁻¹⁷ JAnswer: (d)
Q.8
) In the following nuclear fusion reaction the repalsive potential energy between the two fusing nucler is 7.7× 10⁻¹⁴ J The Temperature to which the gas must be heated is nearly (Boltzman constmt K=1.38 × 10⁻²³ J / K )
0%
a) 103K
0%
b) 105K
0%
c) 107K
0%
d) 109K
Explanation
Both the nucleus have same charge hence repel each other, to overcome repulsion and undergo nuclear fussion, equal amount of kinetic energy must be provided in the form of heat energy Kinetic energy of gas at a given temperature is given by following formula E=(3/2)KT By substituting values 7.7× 10⁻¹⁴ = (3/2)1.38 × 10⁻²³ Answer: (d)
Q.9
) An electron Passing through a Potential diffencne of 4.9 v colides with a mercury atom and trnasfer it to the first excited state what is trnasfer it to the first excited state. what is the wave length of Photon corresponding to the franition of mercury atom to its normal state.
0%
a) 2050A
0%
b) 2935A
0%
c)2525A
0%
d)2240A
Explanation
Energy of electron=E=ve E=4.9 × 1.6 × 10⁻¹⁹ J E=hc / λ ∴ λ=hc / EAnswer: (c)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)