When a metal was exposed to light at a frequency of 3.01× 1015 s–1, electrons we

Question

When a metal was exposed to light at a frequency of 3.01× 1015 s–1, electrons were emitted with a kinetic energy of 5.90× 10–19 J. What is the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 7.65× 10–7 J?

Ionization energy is the energy needed to eject an electron from an atom. Compute the ionization energy of a hydrogen atom in its fourth excited state (n = 5).

The frequency of the middle C note on a piano is 261.63 Hz. What is the wavelength of this note in centimeters? The speed of sound in air is 343.06 m/s.

Explanation / Answer

E=hf

f=frequency=3.01x10^15/s

h=planks constant = 6.626 x 10^-34

K.E= kinetic energy

W= thershold energy

E= K.E + W

W= E-KE= 6.626x10^-34 x 3.01x10^15 - 5.90x10^-19 =19.944x10^-19 - 5.9x10^-19 = 14.044x10^-19J

W=hf'

f' = minimum frequency required for ejection of electron.

14.044x10^-19/6.626x10^-34 = 2.12x10^15/s

total energy = 7.65x10^-7 = h f' n

n= no.of electron = 7.65x10^-7 / 6.626x10^-34 x2.12x10^15 = 54459x10^5

2-

Ionization Energy E = E() - E(4) = 0-E(4) = -E(4)

E(4)=- 1312 xZ/n^2 KJ/mol

for H, Z=1, n= 5

E(4) =- 1312x1 /5x5 =- 52.48KJ/mol

Ionization Energy E= + 52.48 KJ/mol

3- f=261Hz , v = 343.60m/s

velocity = frequency x wavelength

343.60 m/s = 261 Hz x wavelength

wavelength = 343.6/261 =1.3164 m = 131.64 cm

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