Ncert Solutions Chemistry Class 11th
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The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states
The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states
λ1 = 589 nm = 589 x 10-9 m
ν1 = c / λ1 = (3 x 108 ms-1) / (589 x 10-9 m) = 5.0934 x 1014 s-1
λ2 = 589.6 nm = 589.6 x 10-9 m
ν2 = c / λ2 = (3 x 108 ms-1) / (589.6 x 10-9 m) = 5.0882 x 1014 s-1
ΔE = E1 – E2 = h [ν1 – ν2]
= (6.626 x 10-34Js) x [ (5.0934 x 1014 s-1) – (5.0882 x 1014 s-1)
= 3.31 x 10-22 J
Upvote 
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
Time duration, t = 2 ns = 2 x 10-9 s
Frequency, ν = 1 / t = 1 / 2 x 10-9 s = 109 / 2 s-1
Energy of one photon, E = hν = 6.626 x 10-34Js) x (109 / 2 s-1) = 3.25 x 10-25 J
No. of photons = 2.5 x 105
Energy of source = 3.3125 x 10-25 J x 2.5 x 1015 = 8.28 x 10-10 J
Upvote In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
Energy of one photon, E = hν = 6.626 x 10-34Js) x (3 x 108 / 2 ms-1) / 600 x 10-9
= 3.31 x 10-19 J
No. of photons = (3.15 x 10-18) / 3.31 x 10-19 = 9.52 ≈ 10.
Upvote Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate:
(a) Frequency of emission (b) The distance travelled by this radiation in 30s (c) Energy of quantum (d) Number of quanta present if it produces 2 J of energy.
Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate:
(a) Frequency of emission (b) The distance travelled by this radiation in 30s (c) Energy of quantum (d) Number of quanta present if it produces 2 J of energy.
(a) Frequency of emission, ν = c/λ = (3.0 x 108 ms-1) / (616 x 10-9 m) = 4.87 x 1014 s-1
(b) Speed of radiation, c = 3 x 108 ms-1
Distance travelled by this radiation in 30s = 3 x 108 ms-1 x 30 s = 9.0 x 109 m
(c) Energy of quantum, E = hν =hc/λ = [ (6.626 x 10-34Js) x (3 x
Upvote Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.
Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.
This is a short answer type question as classified in NCERT Exemplar
If arenes would undergo electrophilic addition reaction then they would lose their aromaticity which would lead to comparatively less stability. However, alkenes undergo electrophilic addition reaction via carbocation intermediate.
Upvote Nitrogen laser produces radiation of wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 1024, calculate the power of this laser
Nitrogen laser produces radiation of wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 1024, calculate the power of this laser
Power of the laser E = Nhv = Nh c/λ, where N is the number of photos emitted
= [ (5.6 x 1024) x (6.626 x 10-34Js) x (3 x 108 ms-1)] / (337.1 x 10-9 m)
= 3.3 x 106 J
Upvote Arrange the following type of radiations in increasing order of wavelength:
(a) Radiation from microwave oven
(b) Amber light from traffic signal
(c) Radiation from FM radio
(d) Cosmic rays from outer space and
(e) X-rays
Arrange the following type of radiations in increasing order of wavelength:
(a) Radiation from microwave oven
(b) Amber light from traffic signal
(c) Radiation from FM radio
(d) Cosmic rays from outer space and
(e) X-rays
The given radiations in increasing order of wavelength are:
Cosmic rays < X-rays < radiation from microwave oven < amber light from traffic signal < radiation from FM radio
Upvote In the presence of peroxide addition of HBr to propene takes place according to anti Markovnikov’s rule but peroxide effect is not seen in the case of HCl and HI. Explain.
In the presence of peroxide addition of HBr to propene takes place according to anti Markovnikov’s rule but peroxide effect is not seen in the case of HCl and HI. Explain.
This is a long answer type question as classified in NCERT Exemplar
The bond energy of HCl is higher than that of HBr thus it is not cleaved by free radical mechanism to exhibit peroxide effect. However in case of HI the bond energy is so low that the iodine radical forms readily and after formation
Upvote An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Let the no. of electrons in the ion= x
∴ the no. of the protons= x + 3 (as the ion has three units positive charge)
and the no. of neutrons= x + (x×30.4 / 100) = x+ 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons = (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x)
=>2.304x = 56 –
Upvote