Physics Atoms

Get insights from 74 questions on Physics Atoms, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Atoms

Follow Ask Question
74

Questions

0

Discussions

6

Active Users

0

Followers

New answer posted

2 months ago

0 Follower 1 View

S
Syed Aquib Ur Rahman

Contributor-Level 10

Every element has a unique set of spectral lines as its electrons occupy specific energy levels. What scientists know for certain is that these unique patterns act like fingerprints. It helps in identifying elements in stars, flames, or unknown samples. For instance, Helium was discovered in the Sun's spectrum before it was found on Earth.

New answer posted

2 months ago

0 Follower 3 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

We see a discrete emission spectrum when electrons inside excited atoms or ions in gases fall back from higher energy levels to lower ones. Every transition releases a photon of a specific wavelength. Usually the spectrum appears as sharp, bright lines. Dark gaps separate them. These lines are unique to each element.

On the other hand, a continuous emission spectrum is produced when hot solids, liquids, or dense gases emit radiation. This happens because of the collective motion of their atoms and electrons. We don't see sharp lines. Instead the spectrum shows a smooth and unbroken spread of all wavelengths. 

New answer posted

2 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

From the given relations:
mv²/r = |dU/dr| = 4|U? |r³ . (1)
mvr = nh / 2π . (2)
By combining equations (1) and (2), we can derive the radius r:
r = ( (nh)² / (4π√* (U? )*) )¹/³ ⇒ r ∝ n²/³

New answer posted

2 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

E ∝ m ⇒ E / 13.6 = 207 ⇒ E = 207 * 13.6eV = 2815.2eV

New answer posted

2 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

The Balmer series lies in the visible region of the electromagnetic spectrum.

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Velocity of an electron in the nth orbit:

v ∝ 1/n

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

E_ground (H-atom) = -13.6 eV

E_carbon = E_ground (H-atom) * Z²/n²
For carbon, Z=6.

-13.6 eV = -13.6 eV * (6²/n²)

1 = 36/n²

n² = 36

n = 6

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Binding energy = (50M? + 70M? – Msn)C²
= (50.3915 + 70.6069 – 119.902199)U²
= (1.0962U)C²
= 931 * 1.0962MeV

Binding energy per neucleon
= (931 x 1.0962)/120 MeV
= 8.5MeV

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Sol. eVslop = hc/λ - φ

Slope of curve

tan θ = hc/e = constant
as intensity of incident radiation is increased, there will be no effect on graph

New answer posted

2 months ago

0 Follower 11 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

In Rutherford's model, electrostatic force provides the centripetal force:

1 4 πE    

Kinetic Energy (K):

K = 1 2 m v 2 = e 2 8 π E 0 r

Potential Energy (U):

U = 1 4 π E 0 e 2 r  

Total Energy (E):

E = K + U = e 2 8 π E 0 r

 

Note that, E is the Epsilon symbol. 

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

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.