Class 11th
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New answer posted
5 months agoContributor-Level 10
Energy of a hydrogen present in a particular energy shell,
En = 13.12 x 105 / n2 J mol-1 = (13.12 x 105) / (n2 x 6. 022 x 1023) J atom-1
= -2.18 x 10-18 / n2 J atom-1
Step I: Ionisation energy for hydrogen electron present in orbit n = 5
IE5 = E∞ - E5 = 0 – [ (-2.18 x 10-18) / 25] J atom-1 = 8.72 x 10-20 J atom-1
Step II: Ionisation energy for hydrogen electron present in orbit n = 1
IE1 = E∞ - E1 = 0 – [ (-2.18 x 10-18) / 1] J atom-1 = 2.18 x 10-18 J atom-1
Therefore IE1 / IE5 = (2.18 x 10-18 J atom-1) / (8.72 x 10-20 J atom-1) = 25
The energy required to remove an electron from first orbit in a hydrogen atom is 25 times the energy n
New answer posted
5 months agoContributor-Level 10
According to Balmer formula,
Wave number (? ) = RH [1/n12 - 1/n22 ]cm-1
= 109678 [1/22 – 1/42] cm-1
= (109678 x 3) / 16 cm-1
λ = 1 /? = 16 / (109678 x 3) cm = 16 x 107 / (109678 x 3) nm = 486 nm
New answer posted
5 months agoContributor-Level 10
Threshold frequency (v0) = c /λ = (3 * 108 m s-1) / (68 x 10-8 m) = 4.41 x 1014 s-1
Work function (W0) = hv0 = (6.626 * 10-34 Js) x (4.41 x 1014 s-1) = 2.92 x 10-19 J.
New answer posted
5 months agoContributor-Level 10
Energy of one photon (E) = hc / λ
= (6.626 * 10-34 Js) x (3 * 108 m s-1) / (0.57 * 10-6 m) = 3.48 x 10-19 J
Rate of emission of quanta per second = Power / Energy = (25 watt) / (3.48 x 10-19 J)
= (25 Js-1) / (3.48 x 10-19 J) = 7.18 x 1019 s-1
New answer posted
5 months agoContributor-Level 10
Given: λ = 242 nm = 142 x 10-9 m, c = 3 x 108 ms-1, h = 6.626 x 10-34Js
We know, E = hc / λ
= (6.626 * 10-34 Js) x (3 * 108 m s-1) / (242 * 10-9 m) = 0.0821 x 10-17 J
∴ Ionization energy per mol (E) = (0.0821 x 10-17 J) x (6.022 x 1023 mol-1J) / 1000 = 494 kJ mol-1
New answer posted
5 months agoContributor-Level 10
(i) Energy of photon (E) = hc / λ
= (6.626 * 10-34 Js) x (3 * 108 m s-1) / (4 * 10-7 m) = 4.969 x 10-19 J
Since, 1.6020 * 10-19 J= 1 eV
So, 1 J= (1 eV) / (1.6020 * 10-19 J)
Hence, 4.969 x 10-19 J = (1eV) x (4.969 x 10-19 J) / (1.602 x 10-19 J) = 3.1 eV
(ii) Kinetic energy of emission = Energy – work function
= (3.1 – 2.13) = 0.97 eV
(iii) Kinetic energy of emission = 0.97 eV
=> ½ mv2= 0.97 eV = 0.97 x 1.602 x 10-19 J = 0.97 x 1.602 x 10-19 kg m2 s-2
=> v2 = (2 x 0.97 x 1.602 x 10-19 kg m2 s-2 ) / (9.1 x 10-31 kg) = 0.34 x 1012 m2 s-2
=> v = (0.34 x 1012 m2 s-2)1/2 = 0.583 x 106 ms-1 = 5.83 x 105 ms-1
New answer posted
5 months agoContributor-Level 10
Given: h = 6.626 * 10-34 Js,
c = 3 * 108 m s-1,
λ = 4000 pm = 4000 * 10-12 m = 4 * 10-9 m
Energy of photon (E) = hc / λ
= (6.626 * 10-34 Js) x (3 * 108 m s-1) / (4 * 10-9 m) = 4.969 x 10-17 J
i.e. 4.969 x 10-17 J is the energy of 1 photon
Therefore, 1 J is the energy of photons = 1 / (4.969 x 10-17) = 2.012 x 1016 photons.
New answer posted
5 months agoContributor-Level 10
Wavelength, λ = (3 * 108 ms-1) / (5 * 109 s-1) = 6.0 x 10-2 m
Frequency, ν = 1 / 2.0 * 10-10 s = 5.0 x 109 s-1
Wavenumber, (? ) = 1 / (6.0 x 10-2 m) = 16.66 m-1
New answer posted
5 months agoContributor-Level 10
(i) Energy of photon (E) = hν, where h= Plank's const, ν= Frequency
h= 6.626 * 10-34 J s ; ν = 3 * 1015 Hz = 3 * 1015 s-1
∴ E = (6.626 * 10-34 J s) * (3 * 1015 s-1) = 1.986 * 1018 J
(ii) Energy of photon (E) = hν = hc/λ, where λ= wavelength
h = 6.626 * 10-34 J s ; c = 3 * 108 m s-1 λ= 0.50 Å = 0.5 * 10-10 m.
∴ E = (6.626 * 10-34 J s) * (3 * 108 ms-1) / 0.5 * 10-10 m = 3.98 x 10-15 J.
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