Atoms & Nuclei

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149 of 239

Question 149 of 239

The radiation corresponding to 3→23\rightarrow 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3×10−4T3\times 1{0}^{-4}T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

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Answer: (B) 1.1 eV

We know that

R=mυBq\text{R}=\dfrac{\text{m}υ}{\text{Bq}}

∴υ=Bqrm∴υ=\dfrac{\text{Bqr}}{\text{m}}

∴KE=12mv2∴\text{KE}=\dfrac{1}{2}\text{m}{\text{v}}^{2}

=B2q2R22m=\dfrac{{\text{B}}^{2}{\text{q}}^{2}{\text{R}}^{2}}{2m}

= 0.79 eV

Now from photoelectric equation

E=ϕ+KE\text{E}=\phi +\text{KE}

∴ϕ=E−KE∴\phi =\text{E}-\text{KE}

=1.89−0.79=\text{1.89}-\text{0.79}

=1.1 eV=\text{1.1 eV}

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