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State Bohr's third postulate for hydrogen (H_(2) atom. Derive Bohr's formula for the wave number. Obtain expressions for longest and shortest wavelength of spectral lines in utaviolet region for hydrogen atom. The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volt. Monochromatic light of wavelength 2200Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joules. [Charge on electron =1.6xx10^(-19)C] |
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Answer» Solution :Bohr' third postulate : when an electron jupms from higher energy level to LOWER energy level, it RADIATES energy in the form of quanta or photons. The energy of the quantum of electromagnetic radiation i.e., the photon emitted is equal to the energy difference of the two states. Let `E_(i)` is the energy of the electron in a hydrogen atom when it is in on orbit with the principal quantum number `n_(i)` and `E_(F)` is its energy in an orbit with principal quantum number `n_(f)` then `E_(i)=-(me^(4))/(8epsilon_(0)^(2)h^(2)n_(i)^(2))` `E_(f)=-(me^(4))/(8epsilon_(0)^(2)h^(2)n_(f)^(2))` Energy radiated when the electron jumps from higher energy level to lower energy level, `E_(i)-E_(f)=(me^(4))/(8epsilon_(0)^(2)h^(2))((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` This energy is emitted in the form of a quantum of radiation with energy `hv`. `:. hv=(me^(4))/(8epsilon_(0)^(2)h^(2))((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` `:. (1)/(lambda)=(me^(4))/(8epsilon_(0)^(2)h^(3)c)((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` `:. (1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` where, `R=(me^(4))/(8epsilon_(0)^(2)h^(3)c)` is called Rydberg's constant. For longest wavelength in ULTRAVIOLET REGION, `(1)/(lambda_(1))=R((1)/(1^(2))-(1)/(2^(2)))` `lambda_(L)=(4)/(3R)` For shortest wavelength in ultraviolet region, `(1)/(lambda_(s))=R((1)/(1^(2))-(1)/(oo^(2)))` `lambda_(S)=(1)/(R )` Numerical: Given : `V_(0)=1.8` volt, `lambfa=2200Å` `K.E._((max))=eV_(0)` `=1.6xx10^(-19)xx1.8=2.88xx10^(-19)J`. |
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