Question

Solve the question attached.
Physics Dual nature of matter .​

Answer 1

✴ Question :

The work function for the following metals is given: Na : 2.75eV K : 2.30eV Mo : 4.17eV Ni : 5.15eV, which of these metals will not give photoelectric emission for a radiation of wavelength 3300A° from He-Cd laser placed 1m away from the photocell? What happens if the laser is brought nearer and placed 50cm away?

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✴ Answer :

✒ Given :

▪ Φ of Na = 2.75eV

▪ Φ of K = 2.30eV

▪ Φ of Mo = 4.17eV

▪ Φ of Ni = 5.15eV

▪ Wavelength of light = 3300A° = 330nm

✒ To Find :

▪ Which of these metals will not give photoelectric emission.

✒ Concept :

✏ Photoelectric emission occurs only when incident energy is more than that of work function.

✒ Calculation :

✳ Energy released by given wavelength :

→ E = hc/$\lambda$

→ E = 1242/330

→ E = 3.76eV

☞ For Na and K → E > Φ

☞ For Mo and Ni → E < Φ

๏ Therefore, Mo and Ni will not give photoelectric emission.

✴ Energy of radiation is unchanged with decrease in distance.

✴ But, Intensity of emission will increases for Na and K.

✴ Still, Mo and K will not show photoemission.

Answer 2

Given :

★Work function of Na = 2.75 eV

★Work function of K = 2.30 eV

★Work function of Mo = 4.17 eV

★Work function of Ni = 5.15 eV

★Wavelength of radiation = 3300 Å

To Find:

Metals which would not show photoelectric emission for the given wavelength.

Formula Used:

E = hc/λ

Here , E is energy of photon,c(constant) and λ is the wavelength.

Concept:

The energy of photoelectric emission should be always more than the work function i.e E > work function.

Solution:

E = hc/λ

E = 12400/3300 eV

E = 3.75 eV

Now,on comparing the energies

For Na ; E >2.75

For K ; E >2.30

And,

For Mo ; E < 4.17

For Ni ; E < 5.15

Hence, Mo and Ni are the metals that will not give photoelectric emission for the given wavelength of 3300Å.

And, in photoelectric emission decrease in the distance cause no change in energy.Therefore,when the laser is brought nearer then the energy remains unchanged.