Class 11th
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New answer posted
7 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
7 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
7 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
7 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.
New question posted
7 months agoNew question posted
7 months agoNew answer posted
7 months agoContributor-Level 10
(i) One mole of methane (CH4) has = 6.022 * 1023 molecules
No. of electrons present in one molecule of CH4 = 6 + (4 x 1) = 10
No. of electrons present in 6.022 * 1023 molecules of CH4 = 6.022 * 1023 * 10
= 6.022 * 1024 electrons
(ii) (a)
Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 * 103 mg
14 g or 14 * 103 mg of carbon (C-14) has 6.022 * 1023 ato
New answer posted
7 months agoContributor-Level 10
43. A process which can take place of its own or initiate under some conditions. For example: Common salt dissolves in water of its own.
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