A certain shade of blue has a frequency of 7.35 × 1014 Hz. What is the energy of

Question

A certain shade of blue has a frequency of 7.35 × 1014 Hz. What is the energy of exactly one photon of this light? The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, 0.

Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 5.72 × 1014 s–1.

With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of = 225 nm?

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n = 3 to n = 1. To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object.

What kinetic energy must the electrons have in order to resolve a protein molecule that is 4.30 nm in diameter? Take the mass of an electron to be 9.11× 10–31 kg.

Explanation / Answer

a) from E= hf, where h= planck's constant , 6.627*10-34 J.s and f= frequency=7.35*1014 Hz

E= 6.627*10-34*7.35*1014 =48.71*10-20 J, Energy of one photon

b) given f= 5.72*1014/s E= 6.627*10-34*5.72*1014 =37.91*10-20 J

c) from velocity of light, C= f* wave length

frequency = 3*108/225*10-9 =1.333*1015 /s, from E= hf , E= 6.627*10-34*1.333*1015=8.833*10-19 J

kinetic energy of electrons ejected = 8.833*10-19- 3.791*10-19 = 5.042*10-19 J

d) energy of photon = Eo[1 / (n1)2 - 1/(n2)2]

Eo= 13.6 eV and 1ev= 1.6022*10-19 J

E0 = 13.6*(1/1 -1/32)=12.08 eV= 12.08*1.6022*10-19 J=19.35*10-19 J

e) velocity of electrons = h/m*wave length = 6.627*10-34/9.11*10-31*4.3*10-9 =169172.644 m/s

Kinetic energy = 1/2 mV2= 9.11*10-31* (169172.644)2 =2.61*10-20 J

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