Class 11th
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New answer posted
5 months agoContributor-Level 10
This is a short answer type question as classified in NCERT Exemplar
The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1-20 KJ mol-1.
New answer posted
5 months agoContributor-Level 10
λ = 150 pm, v = 1.5 x 107 m s-1
Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10-31 kg x (1.5 x 107 ms-1)2
= [ (9.1 x 1.5 x 1.5) / 2] x 10-31 +14
= 10.2375 x 10-17 J = 1.02375 x 10-16 J
K.E = hc/ λ = (6.626 x 10-34 kg m2 s-1) / (3 x 108 ms-1) / (1.5 x 10-10 m)
= [ (6.626 x 3) x 10-34+8+10] / 1.5
= 13.252 x 10-16 J
We know, E = W0 + K.E.
W0 = E – K.E. = (13.252 – 1.024) x 10-16 J
= 12.228 x 10-16 J
= 12.228 10-16 / 1.602 x 10-19
= 7.63 x 103 eV
New answer posted
5 months agoContributor-Level 10
λ = 256.7 nm = 256.7 x 10-9 m
K.E. = 0.35 eV
E = hc/ λ = (6.626 x 10-34Js) / (3 x 108 ms-1) / (256.7 x 10-9 m)
= (6.626 x 3 x 10-17) J/ 256.7
= (6.626 x 3 x 10-17) / (256.7 x 1.602 x 10-19) eV
E = 4.83 eV
The potential applied to silver gets converted into kinetic energy of the photoelectron.
So, Kinetic energy, K.E= 0.35 V
=> K.E= 0.35 eV
E = W0 + K.E.
=> W0 = E – K.E.
= 4.83 eV – 0.35 eV = 4.48 eV.
New answer posted
5 months agoContributor-Level 10
This is a short answer type question as classified in NCERT Exemplar
Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.
New answer posted
5 months agoContributor-Level 10
Let the threshold wavelength be λ0 nm or λ0 * 10−9 m.
h (ν−ν0) = ½ mv2
hc (1/λ−1/λ0)= ½ mv2
hc [ (1/500*10−9) – (1/ λ0*10−9)] = ½ m (2.55*106)2 . (1)
Similarly,
hc [ (1/450*10−9) – (1/λ0? *10−9? )] = ½? m (4.35*106)2 . (2)
Similarly,
hc [ (1/400*10−9) – (1/λ0? * 10−9? )] = ½? m (5.2 * 106)2 . (3)
Divide equation (2) by (1),
[ (λ0 – 450) /450λ0] x – [500λ0/ (λ0 – 500)] = (4.35/ 2.55)2
(λ0 – 450)/ (λ0 – 500) = 2.61
λ0= 531 nm.
This is the threshold wavelength.
The value of the threshold wavelength is substituted in equation (3).
h *3*108 (1/400*10−9– 1/531*10−9)= ½ *9.1
New answer posted
5 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
5 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
5 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
5 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.
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