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  • The radiation, emitted when an electron jumps from to in a Lit

The radiation, emitted when an electron jumps from\mathrm{n}=4 to \mathrm{n}=3 in a Lithium atom, falls on a metal surface to produce photoelectron. When the photoelectrons with maximum K.E. are made to move perpendicular to a uniform magnetic field of 4 \times 10^{-4} \mathrm{T}, it trace out a circular path of radius 1.68 \mathrm{~cm}. The kinetic energy of electron is
 

Option: 1

3.96 \mathrm{eV}


Option: 2

1.36 \mathrm{eV}


Option: 3

6.8 \mathrm{eV}


Option: 4

2.55 \mathrm{eV}


Answers (1)

best_answer

Radius of circular path of the electron in the magnetic field

\begin{aligned} & \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}} \\ & \mathrm{v}=\frac{\mathrm{qBr}}{\mathrm{m}}=1.18 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}

Maximum KE of the electron = \frac{1}{2} \mathrm{mv}^2=3.96 \mathrm{eV}

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Shailly goel

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