Ncert Solutions Chemistry Class 11th
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Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
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] =
Upvote 
(i) The energy associated with first orbit in hydrogen atom is –2.17 × 10-18 J atom-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
(i) The energy associated with first orbit in hydrogen atom is –2.17 × 10-18 J atom-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
(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
Upvote What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state?
What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state?
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:
Upvote How much energy is required to ionise a hydrogen atom if an electron occupies n = 5 orbit? Compare your answer with the ionisation energy of H atom (energy required to remove the electron from n = 1 orbit)
How much energy is required to ionise a hydrogen atom if an electron occupies n = 5 orbit? Compare your answer with the ionisation energy of H atom (energy required to remove the electron from n = 1 orbit)
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-
Upvote What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n = 4 to energy level n = 2? What is the colour corresponding to this wavelength? (Given RH = 109678 cm-1)
What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n = 4 to energy level n = 2? What is the colour corresponding to this wavelength? (Given RH = 109678 cm-1)
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
Upvote Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0) and work function (W0) of the metal.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0) and work function (W0) of the metal.
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.
Upvote A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.
A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.
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
Upvote Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
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
Upvote A photon of wavelength 4 × 10-7 m strikes on metal surface; the work function of the metal being 2.13 eV. Calculate
(i) The energy of the photon,
(ii) The kinetic energy of the emission
(iii) The velocity of the photoelectron. (Given 1 eV = 1.6020 × 10-19 J).
A photon of wavelength 4 × 10-7 m strikes on metal surface; the work function of the metal being 2.13 eV. Calculate
(i) The energy of the photon,
(ii) The kinetic energy of the emission
(iii) The velocity of the photoelectron. (Given 1 eV = 1.6020 × 10-19 J).
(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 ener
Upvote What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy?
What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy?
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
Upvote