**A photon of wavelength 4 * 10^-7 m strikes on metal surface; the work function of the metal being 2.13 eV. Calculate

(i) The energy of the photon,

(ii) The kinetic energy of the emission

(iii)** The velocity of the photoelectron. (Given 1 eV = 1.6020 * 10^-19 J).

3 Views|Posted 9 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

9 months ago

(i) Energy of photon (E) = hc / λ

= (6.626 * 10^-34 Js) x (3 * 10⁸ m s^-1) / (4 * 10^-7 m) = 4.969 x 10^-19 J

Since, 1.6020 * 10^-19 J= 1 eV

So, 1 J= (1 eV) / (1.6020 * 10^-19 J)

Hence, 4.969 x 10^-19 J = (1eV) x (4.969 x 10^-19 J) / (1.602 x 10^-19J) = 3.1 eV

(ii) Kinetic energy of emission = Energy – work

Upvote

Similar Questions for you

Find out the number of following orders which are INCORRECT against the mentioned properties:

(i) Ortho hydrogen > Para hydrogen (Stability at low temperature)

(ii) Ethanol > Glycerol (Viscosity)

(iii) D₂ < He (Boiling point)

(iv) HF > H₂O (Melting point)

(v) H₃BO₃ > BF₃ (Melting point)

(vi) NH₃ < SbH₃ (Boiling point)

(vii) H₂O₂ > H₂O (Strength of hydrogen bond)

(viii) H₂O < D₂O (Freezing point)

(ix) $H F_{2}^{Θ}$ < HF (Strength of hydrogen bond)

(x) D₂O < H₂O (Bond energy)

Vishal Baghel

Kindly go through the answers

(7.00)

Consider the following reaction:
(A substituted phenol reacts with H? and CrO?Cl to produce product 'P'.)
The product ‘P’ gives positive acetyl chloride test. This is because of the presence of which of these –OH group(s)?


Raj Pandey

Kindly consider the following Image

The number of radial land angular nodes in 4d orbital are, respectively

Payal Gupta

In 4d orbital, n = 4 and $l = 2$

Radial nodes = $n - l - 1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes, $l = 2$

The spin-only magnetic moment value of species is________× 10^-2 BM. (Nearest integer)

[Given:  $\sqrt[]{3} = 1.73]$

Alok Kumar Singh

$B_{2}^{+} \to σ 1 s^{2} s * 1 s^{2} σ 2 s^{2} σ * 2 s^{2} π 2 p^{1}$

Here, number of unpaired electrons, n = 1

Spin only moment ; $μ = \sqrt[]{n (n + 2)} B . M$

= 173 × 10^-2 B.M

Vishal Baghel

$Δ H = Δ U + Δ (P V)$

= $Δ U + P Δ V + V Δ P$

= $Δ U + P Δ V$ (At constant pressure)

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Chemistry Structure of Atom 2025

View Exam Details

2 Answered Questions

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering

A photon of wavelength 4 * 10-7 m strikes on metal surface; the work function of the metal being 2.13 eV. Calculate (i) The energy of the photon, (ii) The kinetic energy of the emission (iii) The velocity of the photoelectron. (Given 1 eV = 1.6020 * 10-19 J).