Question

The photons corresponding to first line of Balmer series of He" ion are incident on Platinum metal which has
work-function of 6.55 eV. Then, which of the following is/are correct?
(A) Platinum does not emit photoelectrons.
(B) Photoelectric effect is observed.
(C) The KE of fastest moving photoelectrons is about 1 ev.
(D) The minimum de-Broglie wavelength of emitted photoelectrons is about 12.3 À​

Answer 1

Answer:

c part is the correct answer

Answer 2

Given:

The photons corresponding to the first line of Balmer series of He⁺ ion are incident on a platinum metal

The work function of Platinum = ∅ = 6.55eV

To Find:

The correct option/s:

(A) Platinum does not emit photoelectrons.

(B) Photoelectric effect is observed.

(C) The KE of fastest moving photoelectrons is about 1 eV.

(D) The minimum de-Broglie wavelength of emitted photoelectrons is about 12.3 A°

Solution:

Statements (B), (C), and (D) are correct.

The first line of the Balmer series has a transition from n₂ = 3 to n₁=2.

The energy of a photon corresponding to the first line of the Balmer series of He⁺ ion = hcRZ² ($\frac{1}{n_1^2} - \frac{1}{n_2^2}$)

= 13.6 X 4 X ( $\frac{1}{4} - \frac{1}{9}$) eV

= 7.39 eV

Since the energy of photon > Threshold energy of Platinum,

⇒ Platinum will eject electrons i.e. photoelectric effect is observed.

Using Einstien's equation,

The KE of moving electrons = E - ∅

= 7.39 - 6.55 eV

= 0.84 eV

⇒ The KE of fastest moving photoelectrons is 0.84 eV ≈ 1eV.

The de-Broglie wavelength corresponding to photoelectrons having KE of 0.84eV = h / mv

= $\frac{h}{\sqrt{2mKE} }$

Here m is the mass of the ejected electron = 9.1 X 10⁻³¹ Kg

= $\frac{12.27}{\sqrt{V} }$ A°

= 12.27 / 1

= 12.27 A° ≈ 12.3 A°

⇒ The minimum de-Broglie wavelength of emitted photoelectrons is about 12.3A°.