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Standard XII

Physics

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When protons ...

Question

When protons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_{A}$ expressed in eV and de-Broglie wavelength $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is $T_{B}$ = ( $T_{A}$ – 1.50 eV). If the de-Broglie wavelength of these photoelectrons is $λ_{B} = 2 λ_{A}$ , then

Work function of A is 2.25 eV

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Work function of B is 4.20 eV

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T_{A}

= 2.0 eV

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T_{B}

= 2.75 eV

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Solution

The correct options are
A Work function of A is 2.25 eV
B Work function of B is 4.20 eV
C $T_{A}$ = 2.0 eV
$K_{m a x} = E - W T_{A} = 4.25 - W_{A} G i v e n T_{B} = T_{A} - 1.50 = 4.70 - W_{B} W_{B} - W_{A} = 1.15 e V λ \propto \frac{1}{\sqrt{K}}, \frac{λ_{B}}{λ_{A}} = \sqrt{\sqrt{\frac{K_{A}}{K_{B}}}} 2 = \sqrt{\frac{T_{A}}{T_{A} - 1.5}} \Rightarrow T_{A} = 2 e V W_{A} = 2.25 e V, W_{B} = 4.20 e V, T_{B} = 0.50 e V$

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Similar questions

Q. When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy,

T_{A}

expressed in eV and de-Broglie wavelength

λ_{A}

. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is

T_{B} = (T_{A} - 1.50 e V)

. If the de-Broglie wavelength of these photoelectrons is

λ_{B} = 2 λ_{A}

, then

Q. When photoelectrons of energy

4.25 e V

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, the ejected photoelectrons have maximum kinetic energy

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and de-Broglie wavelength

λ_{A}

. The maximum kinetic energy of photoelectrons liberated from another metal

B

by photons of energy

4.70 e V

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. If de-Broglie wavelength of these photoelectrons is

λ_{B} = 2 λ_{A}

, then

Q. When photons of energy

4.25 e V

strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy,

T_{A}

e V

and deBroglie wavelength

λ_{A}

. The maximum kinetic energy of photoelectrons liberated from another metal

B

by photons of energy

4.70 e V

T_{B}

(T_{A} - 1.50) e V

. If the de Broglie wavelength of these photoelectrons is

λ_{B} = 2 λ_{A}

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Q. When photons of energy 4.25 eV strike the surface of a metal

A

. The ejected photoelectrons have a maximum kinetic energy

T_{(A)}

(expressed in eV) and de-Broglie wavelength

λ_{(A)}

. The maximum kinetic energy of photoelectrons liberated from another metal

B

by photons of energy 4.2 eV is

T_{B}

. Where

T_{B} = (T_{A} - 1.5)

. If the de-Broglie wavelength of these photoelectrons is

λ_{B}

, where,

(λ_{B} = 2 λ_{A})

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Q. When photons of energy

4.25 e V

strike the surface of a metal

A

, the ejected photo-electrons have maximum kinetic energy,

T_{A}

(expresses in

e V

) and de Broglie wavelength

λ_{A}

. The maximum kinetic energy of photoelectrons liberated from another metal

B

by photons of energy

4.70 e V

T_{B} = T_{A} - 1.50 e V

. If the de-Broglie wavelength of these photo-electrons is

λ_{B} = 2 λ_{A}

, then which is not correct?

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The correct options are A Work function of A is 2.25 eV B Work function of B is 4.20 eV C TA = 2.0 eVKmax=E−WTA=4.25−WAGiven TB=TA−1.50=4.70−WBWB−WA=1.15eVλ∝1√K,λBλA=√√KAKB2=√TATA−1.5⇒TA=2eVWA=2.25eV,WB=4.20eV,TB=0.50eV