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  • Radiation falls on a target within a solenoid of 20 turns per cm. carrying a current 2.5 amps. Electrons emitted move in a circle of largest radius 1 cm. Then the wavelength of radiation given the

Radiation falls on a target within a solenoid of 20 turns per cm. carrying a current 2.5 amps. Electrons emitted move in a circle of largest radius 1 cm. Then the wavelength of radiation given the work function of the target is 0.5 volts,

e=1.6 \times 10^{-19} \text { coulomb, } \mathrm{h}=6.625 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{m}=9.1 \times 10^{-31} \mathrm{Kg}

Option: 1

35.24\, \AA


Option: 2

41.25\, \AA


Option: 3

10.25 \, \AA


Option: 4

22.45\, \AA


Answers (1)

best_answer

Magnetic flux density inside the solenoid

\mathrm{B}=\mu_0 \mathrm{ni}=\left(4 \pi \times 10^{-7}\right) \times(20 \times 100) \times 2.5=2 \pi \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^2

The ejected electrons move in a circle under the influence of magnetic field. If v be the velocity of ejected electron, then \left(m v^2 / r\right)=B e v

\text { or } \quad v=\frac{\text { Ber }}{m}=\frac{2 \pi \times 10^{-3} \times 1.6 \times 10^{-19} \times 10^{-2}}{9.1 \times 10^{-31}}=11.12 \times 10^6 \mathrm{~m} / \mathrm{sec}

Now, according to Einstein’s photoelectric equation,

\mathrm{hv}=(1 / 2) \mathrm{mv}^2+\mathrm{W} \text { (Where } \mathrm{W}=\text { Work function) }

=(1 / 2) \times\left(9.1 \times 10^{-31}\right) \times\left(11.12 \times 10^6\right)^2+0.5 \times 1.6 \times 10^{-19}=564 \times 10^{-19}

\therefore \quad v=\frac{564 \times 10^{-19}}{6.625 \times 10^{-34}}=85.13 \times 10^{15}

\text { Now, } \lambda=\frac{\mathrm{c}}{{v}}=\frac{3 \times 10^8}{85.13 \times 10^{15}}

=3.524 \times 10^{-9} \mathrm{~m}=35.24 \AA \text {. }

Posted by

Gautam harsolia

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