Dual Nature of Radiation and Matter

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3 months ago

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A
alok kumar singh

Contributor-Level 10

Using de-Broglie equation:
λ = h/p = h/√ (2mK) = h/√ (2meV)
λ ∝ 1/√m
λ? /λ? = √ (m? /m? ) = √ (1.00727/0.00055) = √1831 ≈ 43:1

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

K = hc/λ - W?
K = (20*10? ²? )/ (500*10? ) - 1.25 * 1.6*10? ¹?
K = 4*10? ¹? - 2*10? ¹? = 2*10? ¹? J
R = √ (2mK)/ (eB)
B = √ (2mK)/ (eR) = √ (2 * 9*10? ³¹ * 2*10? ¹? ) / (1.6*10? ¹? * 30*10? ²)
B = √ (36*10? ) / (48*10? ²¹) = (6*10? ²? ) / (4.8*10? ²? ) = 1.25 * 10? T
B = 125 * 10? T

New answer posted

3 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

For photon, λ? = h/p = 6.6*10? ³? /10? ²? = 6.6*10? m
For a particle, λ? = h/mv = 6.6*10? ³? / (9.1*10? ³¹ * 10? ) ≈ 7.25*10? ¹? m
λ? /λ? = 6.6*10? / 7.25*10? ¹? ≈ 910

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

MV? = 3MV? = p
λ = h/mv = h/p
λ? /λ? = 1

New answer posted

3 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

By Einstein's equation of photoelectric effect,
hc/λ = hc/λ? + eVs
Now hc/λ = hc/λ? + e*4.8 . (i)
hc/ (2λ) = hc/λ? + e*1.6 . (ii)
Multiply (ii) by 2:
hc/λ = 2hc/λ? + e*3.2 . (iii)
Equating (i) and (iii):
hc/λ? + 4.8e = 2hc/λ? + 3.2e
1.6e = hc/λ?
Substitute this into (i)
hc/λ = 1.6e + 4.8e = 6.4e
Substitute this into (ii)
(6.4e)/2 = hc/λ? + 1.6e
3.2e = hc/λ? + 1.6e
1.6e = hc/λ?
From (i): hc/λ = hc/λ? + 3 (hc/λ? ) = 4hc/λ?
λ? = 4λ

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

For electron,
λe = h/p ⇒ (KE)e = ½mv² = p²/ (2m) = (h²/λ²)/ (2m)
For photon,
λp = h/p ⇒ (K.E.)p = pc = hc/λ
KEe / KEp = (h²/ (2mλ²) / (hc/λ) = h/ (2mcλ)
But for electron p = mv = h/λ so h/λ = mv
KEe / KEp = mv / (2mc) = v/2c

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

λ = h/mv = h/√2mK
For same K:
λ ∝ 1/√m
λ? : λ? : λ? = 1/√m? : 1/√m? : 1/√4m?
As m? > m? ,
λ? > λ? > λ?

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

p = h λ  

So, two photons having equal linear momenta have equal wavelength. As wavelength decreases, momentum and energy of photon increases.

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Case I :- 2 ? ? = 1 2 m v 1 2 . . . . . . . . ( i )

Case II :- 1 0 ? ? = 1 2 m v 2 2 . . . . . . . . ( i i )

1 9 = v 1 2 v 2 2

v 1 : v 2 = 1 : 3

x = 1

New question posted

3 months ago

0 Follower 3 Views

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