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The path of photoelectrons emitted due to electromagnetic radiation incident on a sample of material A, is found to have a maximum bneding radium of 0.1 m in a magnetic field of ((sqrt2)/(3))xx10^-4T. When the radiation is incident normally on a double slit having a slit separation of 0.1mm, it is observed that there are 10 fringes in a width of 3.1 cm on a screen placed at a distance of 1 m from the double slit. Find the work function of the material. and the corresponding threshold wavelength. What sould be the wavelength of the incident light so that the bending radius is one-half of what it was before? Given that mass of the electron=0.5Me(V)/(c^2),hc=12400eV-A |
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Answer» Solution :The bending radias r of a PARTICLE of mass m and charge e moving with a velocity v in a uniform magnetic FIELD is given by `(mv^(2))/r=evB`, e being the charge and B the strength of the firld. Momentum, `mv=BER`..(i) When the em radiation, `lamda`, satisfies, `FW=(lamdaD)/(d)` `lamda=(d)/(D)xxFW` `=(0.1xx10^(-3)xx3.1xx10^-3)/(1)m=3100A` The kinetic energy of the electron is `KE=((Ber)^(2))/(2m)=((Brec)^(2))/(2mc^(2))=(((sqrt2)/(3)xx10^(-4)xx10^(-1)xx3xx10^(8))^(2))/(2xx0.5xx10^(6))=2eV` The threshold wavelength is given by `(hc)/(lamda)=KE+(hc)/(lamda_(th))` `(12400)/(3100)=2+(hc)/(lamda_(th))` `(hc)/(lamda_(th))=2eV`,`lamda_(th)=(12400)/(2)=6200A` The kinetic energy of the most energetic photoelectron is directly proportional to the square of the radius of its track in a magnetic field. So, if the bending radius is half of what it was before, the kinetic energy of the electron is one-fourth of the previous value. `(KE')=2xx((1)/(2))^(2)EV` or `0.5eV` The wavelength, `lamda`, is given by `(hc)/(lamda)=2eV+0.5eV=2.5eV` `implieslamda=(12400)/(2.5)=4960 A` |
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