Light of wavelength $1824 A 0$ , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength $1216 A 0$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

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Answer 1

As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 A o$ ) or $10.17 e V$ (for $λ = 1216 A o$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$

Answer 2

As we know,

E = W_{0} + K_{m a x}

. From the given data

E

is

6.75 e V

(for

λ = 1824 A o

) or

10.17 e V

(for

λ = 1216 A o

)

∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V

OR

W_{0} = 10.17 - 8.7 = 1.5 e V

Light of wavelength $1824 A 0$ , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength $1216 A 0$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

6.8 eV

1.5 eV

13.6 eV

3.5 eV

Solution

As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 A o$ ) or $10.17 e V$ (for $λ = 1216 A o$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$

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Light of wavelength 1824A0 , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength 1216A0 is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

6.8 eV

1.5 eV

13.6 eV

3.5 eV

As we know,E=W0​+Kmax​. From the given data E is 6.75 eV (for λ=1824Ao) or 10.17 eV (for λ=1216Ao) ∴W0​=E−Kmax​=6.75−5.3=1.48 eV OR W0​=10.17−8.7=1.5 eV

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More From Chapter

Dual Nature of Radiation and Matter

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Light of wavelength $1824 A 0$ , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength $1216 A 0$ is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is :

As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 A o$ ) or $10.17 e V$ (for $λ = 1216 A o$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$

The maximum velocity of an electron by light of wavelength $γ$ incident on the surface of a metal of work function $ϕ$ is

Light of wavelength 4000 $\overset{˚}{A}$ is incident on a metal surface. The maximum kinetic energy of emitted photoelectron is 2 eV. What is the work function of the metal surface?

A photon of energy $40$ eV is incident on a surface of metal. Due to this an electron is emitted of kinetic energy $37.5 e V$ . Find out the work function of the metal :

Consider a metal exposed to light of wavelength $600 n m$ . The maximum energy of the electron doubles when light of wavelength $400 n m$ is used. Find the work function in $e V$ .