Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 11.26:

Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼10⁵ W m⁻²) red light of wavelength 6328 Å produced by a He-Ne laser?

Answer:

Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10⁻¹⁰ m

Stopping potential of the metal, V₀ = 1.3 V

Planck’s constant, h = 6.6 × 10⁻³⁴ J

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Work function of the metal = 

Frequency of light = ν

We have the photo-energy relation from the photoelectric effect as:

 = hν − eV₀

Let ν₀ be the threshold frequency of the metal.

∴ = hν₀

Wavelength of red light,   = 6328 × 10⁻¹⁰ m

∴Frequency of red light, 

Since ν₀> νr, the photocell will not respond to the red light produced by the laser.

Question 11.27:

Monochromatic radiation of wavelength 640.2 nm (1nm = 10⁻⁹ m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

Wavelength of the monochromatic radiation, λ = 640.2 nm

= 640.2 × 10⁻⁹ m

Stopping potential of the neon lamp, V₀ = 0.54 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Let   be the work function and ν be the frequency of emitted light.

We have the photo-energy relation from the photoelectric effect as:

eV₀ = hν − 

Wavelength of the radiation emitted from an iron source, λ‘ = 427.2 nm

= 427.2 × 10⁻⁹ m

Let  be the new stopping potential. Hence, photo-energy is given as:

Hence, the new stopping potential is 1.50 eV.

Question 11.28:

A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

λ₁ = 3650 Å, λ₂= 4047 Å, λ₃= 4358 Å, λ₄= 5461 Å, λ₅= 6907 Å,

The stopping voltages, respectively, were measured to be:

V₀₁ = 1.28 V, V₀₂ = 0.95 V, V₀₃ = 0.74 V, V₀₄ = 0.16 V, V₀₅ = 0 V

Determine the value of Planck’s constant h, the threshold frequency and work function for the material.

[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10⁻¹⁹ C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]

Answer:

Einstein’s photoelectric equation is given as:

eV₀ = hν− 

Where,

V₀ = Stopping potential

h = Planck’s constant

e = Charge on an electron

ν = Frequency of radiation

 = Work function of a material

It can be concluded from equation (1) that potential V₀ is directly proportional to frequency ν.

Frequency is also given by the relation:

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

The given quantities can be listed in tabular form as:

The following figure shows a graph between νand V_0.

It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 × 10¹⁴ Hz, which is the threshold frequency (ν₀) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ₅ line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = 

From equation (1), the slope   can be written as:

The work function of the metal is given as:

= hν₀

= 6.573 × 10⁻³⁴ × 5 × 10¹⁴

= 3.286 × 10⁻¹⁹ J

= 2.054 eV

Question 11.29:

The work function for the following metals is given:

Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Answer:

Mo and Ni will not show photoelectric emission in both cases

Wavelength for a radiation, λ = 3300 Å = 3300 × 10⁻¹⁰ m

Speed of light, c = 3 × 10⁸ m/s

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

The energy of incident radiation is given as:

It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

Question 11.30:

Light of intensity 10⁻⁵ W m⁻² falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your Answer?

Answer:

Intensity of incident light, I = 10⁻⁵ W m⁻²

Surface area of a sodium photocell, A = 2 cm² = 2 × 10⁻⁴ m²

Incident power of the light, P = I × A

= 10⁻⁵ × 2 × 10⁻⁴

= 2 × 10⁻⁹ W

Work function of the metal,  = 2 eV

= 2 × 1.6 × 10⁻¹⁹

= 3.2 × 10⁻¹⁹ J

Number of layers of sodium that absorbs the incident energy, n = 5

We know that the effective atomic area of a sodium atom, Ae is 10⁻²⁰ m²_.

Hence, the number of conduction electrons in n layers is given as:

The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:

Time required for photoelectric emission:

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

Question 11.31:

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me= 9.11 × 10⁻³¹ kg).

Answer:

An X-ray probe has a greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe, λ = 1 Å = 10⁻¹⁰ m

Mass of an electron, me = 9.11 × 10⁻³¹ kg

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

The kinetic energy of the electron is given as:

Where,

v = Velocity of the electron

mev = Momentum (p) of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

Energy of a photon, 

Hence, a photon has a greater energy than an electron for the same wavelength.

Question 11.32:

(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 × 10⁻²⁷ kg)

(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Answer:

(a) De Broglie wavelength = ; neutron is not suitable for the diffraction experiment

Kinetic energy of the neutron, K = 150 eV

= 150 × 1.6 × 10⁻¹⁹

= 2.4 × 10⁻¹⁷ J

Mass of a neutron, mn = 1.675 × 10⁻²⁷ kg

The kinetic energy of the neutron is given by the relation:

Where,

v = Velocity of the neutron

mnv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e., 10⁻¹⁰ m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

(b) De Broglie wavelength = 

Room temperature, T = 27°C = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

Where,

k = Boltzmann constant = 1.38 × 10⁻²³ J mol⁻¹ K⁻¹

The wavelength of the neutron is given as:

This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high-energy neutron beam should first be thermalised, before using it for diffraction.