Light of wavelength $0.6 μ m$ from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is $0.5 V$ . With light of wavelength $0.4 μ m$ from a mercury vapor lamp, the stopping potential is $1.5 V$ . Then, the work function [in electron volts] of the photocell surface is

0.75 e V

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1.5 e V

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3 e V

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2.5 e V

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Solution

The correct option is A $1.5 e V$
Energy of $0.6 μ m$ is $\frac{h c}{0.6 * 10^{- 6}}$ $= 2.05 e V$

stopping potential is $0.5 v$ i.e, $K . E = 0.5 e V$
work function of material = $2.05 - 0.5 e V$
= $1.5 e V$

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Similar questions

Light of wavelength $0.6 μ m$ from a sodium lamp falls on a photo cell and causes the emission of photoelectrons for which the stopping potential is 0.5V with light of wavelength $0.4 μ m$ from a sodium lamp, the stopping potential is 1.5V, with this data the value of $\frac{h}{e}$ is:

Q. Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5 V. With this data, the value of h/e is:

Q. Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is

5.0 V

and the work function of sodium is

2.2 e V

. What is the wavelength of the incident light?

Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is 5.0 V, and the work function of sodium is 2.2 eV. What is the wavelength of the incident light? __nm.

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The correct option is A 1.5 eVEnergy of 0.6μm is hc0.6∗10−6 =2.05eVstopping potential is 0.5v i.e, K.E=0.5eVwork function of material = 2.05−0.5eV = 1.5eV

The correct option is A $1.5 e V$
Energy of $0.6 μ m$ is $\frac{h c}{0.6 * 10^{- 6}}$ $= 2.05 e V$

stopping potential is $0.5 v$ i.e, $K . E = 0.5 e V$
work function of material = $2.05 - 0.5 e V$
= $1.5 e V$