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A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of $60^{\circ}$ with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm. Find the work function of the metal in electron volt. Given that specific charge of electron is 1.76 x 10¹¹ C/kg.

A photon with an energy of 4.9 eV ejects photoelectrons from tungsten……

Solution.

Pitch of helical path, $\displaystyle P=\left( v\cos \theta \right)T=\frac{vT}{2}$

(as θ = 60º)

Here $\displaystyle T=\frac{2\pi m}{qB}=\frac{2\pi }{B\alpha }$

$\displaystyle (\alpha =\frac{q}{m})$

Hence Pitch, $\displaystyle P=\frac{\pi v}{B\alpha }$

$\displaystyle \Rightarrow v=\frac{B\alpha P}{\pi }=\frac{(2.5\times {{10}^{-3}})\times (1.76\times {{10}^{11}})\times (2.7\times {{10}^{-3}})}{3.14}$

$\displaystyle \Rightarrow v=0.38\times {{10}^{6}}m{{s}^{-1}}$

Now

E = Φ₀ + K_max

⇒ Φ₀ = E – K_max

$\displaystyle {{\phi }_{0}}=E-\frac{1}{2}m{{v}^{2}}$

$\displaystyle {{\phi }_{0}}=4.9-\frac{1}{2}\frac{(9.1\times {{10}^{-31}})\times {{(0.38\times {{10}^{6}})}^{2}}}{1.6\times {{10}^{-19}}}$

⇒ Φ₀ = (4.9 – 0.4) eV

⇒ Φ₀ = 4.5 eV

A photon with an energy of 4.9eV ejects photoelectrons from tungsten

A photon with an energy of 4.9 eV ejects photoelectrons from tungsten……

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