Question

A photon with wavelength of with strikes a metal surface an electron that has a 2.500 Å and ejects. KE of 1.6 ev i) Find P, ii) Find de-broglie wavelength of the ejected electron iii) Assume this electron is more energetic possible, what is the work function of the metal . iv) What is the max. wavelength of wavelength of photon that will eject electron an from the metal surface.


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Answer 1

$\huge\mathfrak\pink{Answer}$

We know λ=5×10 –7 m(given)

C=3×10 8

From the equation E=hv orhc/λ

Where, h = Planck’s constant =6.626×10 –34Js

c = velocity of light in vacuum =3×10 8m/s

λ= wavelength of photon =5×10

–7 m

Substituting the values in the given expression of E:

E=(6.26×10 −34 )(3×10 8 )/5×10 −7

=3.9695×10 −19 J

Hence, the energy of the photon is 3.97×10 –19 J.

Answer 2

Answer:

We know λ=5×10 –7 m(given)

C=3×10 8

From the equation E=hv orhc/λ

Where, h = Planck’s constant =6.626×10 –34Js

c = velocity of light in vacuum =3×10 8m/s

λ= wavelength of photon =5×10

–7 m

Substituting the values in the given expression of E:

E=(6.26×10 −34 )(3×10 8 )/5×10 −7

=3.9695×10 −19 J

Hence, the energy of the photon is 3.97×10 –19 J.

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Explanation:

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