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.
Given λ = 6800 amgstrom or 6800x 10-10 m = 6.8 x 10-7m
C = 3 x 108 m/s
Also λ = c/v or v = c/ λ Therefore putting the values in the equation, we get
=3 x 108 / 6.8 x 10-7m = 4.41 x 1014 s-1
Hence, work function (W0) of the metal = hν0
= (6.626 × 10–34 Js) (4.41 × 1014 s–1)
= 2.922 × 10–19 J
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(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
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(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?
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(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
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(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
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Thanks
Work function=h*threshold frequency Not h*frequency
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velocity of light
c is what and how 3*10*8