Question

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Answer 1

question:

The kinetic energy of an electron is E when the incident wavelength is λ. To increase the kinetic energy of the electron to 2E, the incident wavelength must be :

answer:

The kinetic energy of an electron is E when the incident wavelength is λ. To increase the kinetic energy of the electron to 2E, the incident wavelength must be :

The Einstein's equation for photoelectric effect is given by , KE=

λ

hc

−W where KE= kinetic energy of electron, h= Planck's constant, W= work function of metal , c= velocity of light and

λ= wavelength of incident light .

When KE=E⇒E=

λ

hc

−W..(1)

and KE=2E⇒2E=

λ

′

hc

−W..(2)

Using (1), (2) becomes, 2E=

λ

′

hc

+E−

λ

hc

or

λ

′

hc

=E+

λ

hc

or λ

′

=

Eλ+hc

hcλ

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Answer 2

Answer:

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Explanation:

question:

The kinetic energy of an electron is E when the incident wavelength is λ. To increase the kinetic energy of the electron to 2E, the incident wavelength must be :

answer:

The kinetic energy of an electron is E when the incident wavelength is λ. To increase the kinetic energy of the electron to 2E, the incident wavelength must be :

The Einstein's equation for photoelectric effect is given by , KE=

λ

hc

−W where KE= kinetic energy of electron, h= Planck's constant, W= work function of metal , c= velocity of light and

λ= wavelength of incident light .

When KE=E⇒E=

λ

hc

−W..(1)

and KE=2E⇒2E=

λ

′

hc

−W..(2)

Using (1), (2) becomes, 2E=

λ

′

hc

+E−

λ

hc

or

λ

′

hc

=E+

λ

hc

or λ

′

=

Eλ+hc

hcλ

please make brainliest