This set of Class 11 Chemistry Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Developments Leading to the Bohr’s Model of Atom”.
1. What’s the wavelength for the visible region in electromagnetic radiation?
a) 400 – 750 nm
b) 400 – 750 mm
c) 400 – 750 μm
d) 400 – 750 pm
Explanation: Electromagnetic spectrum is made up of various electromagnetic radiation. They are radio waves, X-rays, gamma rays, UV rays, the visible region, IR waves, and microwaves. Visible rays are the only ones which a human eye can see. They range from 450 – 750 nm.
2. What is the wavenumber of violet color?
a) 25 x 103 mm-1
b) 25 x 103 m-1
c) 25 x 103 cm-1
d) 25 x 103 nm-1
Explanation: The wavenumber is the reciprocal or the inverse of wavelength. Wavenumber = 1/Wavelength. Its unit is cm-1. The wavelength of violet color is 400nm as seen in the electromagnetic spectrum. So wavenumber = 1/400nm = 25 x 103 cm-1.
3. Calculate the frequency of the wave whose wavelength is 10nm.
a) 2 Hz
b) 3 Hz
c) 1 Hz
d) 4 Hz
Explanation: The relation between wavelength(λ) and frequency(v) of a wave is given by λ = c/v where c is the speed of light of the light. v = c/λ Frequency of the given wave = (3 x 108m/s)/(10 x 10-9m) = 3 Hz.
4. If Energy = 4.5 KJ; calculate the wavelength.
a) 4.42 x 10-29 m
b) 4.42 x 10-39 m
c) 4.42 x 10-25 m
d) 4.42 x 10-22 m
Explanation: We know E = hv through Planck’s Quantum Theory, where E is energy, h is Planck’s constant and v is the frequency. 4.5 KJ = (6.626×10–34 Js)(3 x 108m/s)/(wavelength). wavelength = 4.42 x 10-29 m.
5. ________ frequency, is the minimum frequency required to eject an electron when photons hit the metal surface.
Explanation: In the photoelectric effect, when photons strike on a metal surface, it emits electrons. Thus for emitting an electron, it requires a minimum amount of energy. This is threshold energy acquired through threshold frequency.
6. A metal’s work function is 3.8KJ. Photons strike metal’s surface with an energy of 5.2 KJ. what’s the kinetic energy of the emitted electrons?
a) 3.8 KJ
b) 5.2 KJ
c) 9 KJ
d) 1.4 KJ
Explanation: As per the formula of the photoelectric effect, we have E = K.E. + Wo. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and Wo is the work function. K.E. = 5.2 KJ – 3.8 KJ = 1.4 KJ.
7. When an electron jumps from 3rd orbit to 2nd orbit, which series of spectral lines are obtained?
Explanation: As per the spectral lines of the hydrogen, when an electron jumps from nth orbit to 2nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in visible region.
8. Find out the wavenumber, when an electron jumps from 2nd orbit to 1st.
a) 82357.75 cm-1
b) 105,677 cm-1
c) 82257.75 cm-1
d) 109,677 cm-1
Explanation: The Swedish spectroscopist, Johannes Rydberg gave a formula; Wavenumber = RH[(1/n1)2-(1/n2)2]. Here RH is the Rydberg constant and is equal to 109,677 cm-1. Wavenumber = 109,677(3/4) = 82257.75 cm-1.
9. The ultraviolet spectral region is obtained in Balmer series.
Explanation: When an electron jumps from nth orbit to 1st orbit, provided that n = 1, 2, 3, etc, it emits radion in the ultraviolet region. As per the spectral lines of the hydrogen, when an electron jumps from nth orbit to 2nd orbit, it’s in Balmer series (provided that n = 3, 4, 5….). For Balmer series, the electron emits waves in the visible region.
10. During the photoelectric effect, when photons strike with 5.1eV, electrons emitted from which metal have higher kinetic energy?
|Work function||2.3 eV||4.3 eV|
Explanation: As per the formula of the photoelectric effect, we have E = K.E. + Wo. E is the energy of photons; K.E. is the kinetic energy with which electrons are emitted and Wo is the work function. K.E. of Na and Ag are 2.8 eV and 0.8 eV.
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