Light of wavelength $$2475A0$$ is incident on Barium. Photoelectrons emitted describe a circle of radius $$100$$ cm by a magnetic field of flux density $$117$$×$$10$$−$$5Tesla$$. The workfunction of Barium is (Given$$ $$em=1.7$$×$$1011$$ $$C/Kg)

Question

Light of wavelength $$2475A0$$  is incident on Barium. Photoelectrons emitted describe a circle of radius $$100$$ cm by a magnetic field of flux density $$117$$×$$10$$−$$5Tesla$$. The workfunction of Barium is (Given$$ $$em=1.7$$×$$1011$$ $$C/Kg)

Infinity Learn · Accepted Answer

Radius of circular path described by the charged particle in the given magnetic field is

$\begin{array}{l} r = \frac{\sqrt{2 m (K E)}}{q B} \\ K E = \frac{q^{2} B^{2} r^{2}}{2 m} = \frac{e}{m} \times \frac{e B^{2} r^{2}}{2} \\ = 1.7 \times 10^{11} \times \frac{1}{2} \times 1.6 \times 10^{- 19} \times {(\frac{1}{\sqrt{17}} \times 10^{- 5})}^{2} \times 1^{2} \\ = 8 \times 10^{- 20} J \\ = 0.5 e V \end{array}$

By using the relation,

$W_{0} = E - K E_{\max} = \frac{12400}{2475} - 0.5 = 4.5 e V$

Photo Electric Effect

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Light of wavelength 2475 $\overset{0}{A}$ is incident on Barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density $\frac{1}{\sqrt{17}} \times 10^{- 5}$ Tesla. The workfunction of Barium is (Given $\frac{e}{m} = 1.7 \times 10^{11}$ C/Kg)

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Photo Electric Effect

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Light of wavelength 2475A0 is incident on Barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density 117×10−5Tesla. The workfunction of Barium is (Given em=1.7×1011 C/Kg)

Radius of circular path described by the charged particle in the given magnetic field isr=2mKEqBKE=q2B2r22m=em×eB2r22 =1.7×1011×12×1.6×10−19×117×10−52×12 =8×10−20J =0.5 eVBy using the relation, W0=E−KEmax=124002475−0.5=4.5eV

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Light of wavelength 2475 $\overset{0}{A}$ is incident on Barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density $\frac{1}{\sqrt{17}} \times 10^{- 5}$ Tesla. The workfunction of Barium is (Given $\frac{e}{m} = 1.7 \times 10^{11}$ C/Kg)