Class 11th
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New answer posted
7 months agoContributor-Level 10
W0 = 1.9 eV = 1.9 x 1.602 x 10-19 J
Threshold frequency, v0 = W0 / h = (1.9 x 1.602 x 10-19 J) / (6.626 x 10-34Js)
= 0.459 x 1015 s-1 = 4.59 x 1014 s-1
Threshold wavelength? 0 = c / v0 = (3 x 108 ms-1) / (4.59 x 1014 s-1)
= 0.6536 x 10-6 m = 653.6 nm? 654 nm
Kinetic energy, E = E0 + ½ mv2
(1/2 mv2) = E – E0 = hc [ (1/? ) – (1/? 0)]
= (6.626 x 10-34Js) x (3 x 108 ms-1) / (10-9) x [ (1/500) – (1/654)]
= 6.626 x 3 x 154 x 10-34+8+9) / (500 x 654)
= 9.36 x 10-20 J
Velocity, v = [ (2 x 9.36 x 10-20) / m]1/2
= [ (2 x 9.36 x 10-20) kg m2 s-2 / 9.1 x 10-31 kg]1/2
= (2.057 x 1011 m2s-2)1/2= (20.57 x 1010 m2s-2)1/2
= 4.5356 x 105 ms-1
New answer posted
7 months agoContributor-Level 10
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
New answer posted
7 months agoContributor-Level 10
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
New answer posted
7 months agoContributor-Level 10
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.
New question posted
7 months agoNew answer posted
7 months agoContributor-Level 10
(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 108 ms-1)] / (616 x 10-9 m) = 32.27 x 10-20 J
(d) Number of quanta present if it produces 2 J of energy
&n
New answer posted
7 months agoContributor-Level 10
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.
New answer posted
7 months agoContributor-Level 10
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
New answer posted
7 months agoContributor-Level 10
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
New answer posted
7 months agoContributor-Level 10
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 it combines to form an iodine molecule.
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