Chemistry
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New answer posted
a year agoContributor-Level 10
Step I: calculation of energy required
Energy of electron (En) = (- 2.18 x 10-11ergs) / n2= (- 2.18 x 10-18 J) / n2
Energy in Bohr's 1st orbit (E1) = (- 2.18 x 10-18 J) / 12
Energy in Bohr's 5th orbit (E1) = (- 2.18 x 10-18 J) / 52
Therefore, energy required (ΔE) = E5 – E1 = [ (- 2.18 x 10-18 J) / 25] – [ (- 2.18 x 10-18 J) / 1]
= 2.18 x 10-18 (1 – 1/25) J
= 2.18 x 10-18 x 24 / 25
= 2.09 x 10-18 J
Step II: Calculation of wavelength of light emitted
λ = hc / ΔE = (6.626 x 10-34 J s) x (3 x 108 ms-1) / (2.09 x 10-18 J) = 9.50 x 10-8 m = 950
New answer posted
a year agoContributor-Level 10
According to Balmer formula? = 1 / λ = RH [1/n12 – 1/n22]
For the longest wavelength transition in the Balmer series of atomic hydrogen, wave number must be least. This is possible in case n2 – n1 = minimum; i.e. n1 = 2 and n2 = 3. Substituting the values:
? = 1 / λ = (1.097 x 107 m-1) [1/22 – 1/32] = (1.097 x 107 m-1) [5/36] = 1.523 x 106 m-1
New answer posted
a year agoContributor-Level 10
(i) For an electron, the energies in the two orbits can be compared as:
E1 / E2 = (n2 / n1)2 [Since E is inversely proportional to n2]
Given: n1 = 1, E1 = –2.17 * 10-18 J atom-1, n2 = 5
Therefore, (–2.17 * 10-18 J atom-1) / E2 = (5 / 1) 2 = 25
=> E2 = (–2.17 * 10-18 J atom-1) / 25 = –8.77 x 10-20 J atom-1
(ii) For hydrogen atom; rn = 0.529 x n2 Å
r5 = 0.529 x (5)2 = 13.225 Å = 1.3225 nm.
New answer posted
a year agoContributor-Level 10
The maximum no. of emission lines = [n (n–1)] /2 = [6 (6–1)] / 2 =3 * 5 = 15
The transitions that take place are as follows:
New answer posted
a year 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
a year 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
a year 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
a year 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
a year 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
a year 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
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