Dual Nature of Radiation and Matter
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New answer posted
a month agoContributor-Level 10
In case 1, E? = 2eV; φ? ? = E? - KE? = 1eV
In case 2, E? = 12400/1550 = 8eV; max.KE = 8eV - 1eV = 7eV
New answer posted
a month agoContributor-Level 10
So, two photons having equal linear momenta have equal wavelength. As wavelength decreases, momentum and energy of photon increases.
New answer posted
a month agoContributor-Level 10
λ? = h/√2mE? = λ? = h/√2mE?
=> E? = (4/9)E? = 4eV
E? = E? - eV? => V? = 5V
New answer posted
a month agoContributor-Level 10
p = nhc/λ ⇒ n = pλ/hc
n = (3.3 * 10? ³ * 600 * 10? ) / (6.6 * 10? ³? * 3 * 10? ) = 10¹?
New answer posted
a month agoContributor-Level 10
hc/λ = K? + φ [given φ is negligible]
So, hc/λ = K?
λ? = h/√ (2mK? ²) ⇒ K? = h²/ (2mλ? ²)
(hc)/λ = h²/ (2mλ? ²) ⇒ λ = (2mc/h)λ? ²
New answer posted
a month agoContributor-Level 9
Incident energy
If electron will emit.
No electron emission
Only caesium will emit electron
New answer posted
a month agoContributor-Level 10
R = (2μ sinθ) / (1.22λ) . (1)
According to de-Broglie's hypothesis, we can write
λ = h / (mv) . (2)
With the help of equations (1) and (2), we can write
R = (2μmv sinθ) / (1.22h) ⇒ R? /R? = m? /m? = 1837 ⇒ R? = 1837R?
- Concept involved: Resolving Power of Microscope
- Topic: Optics
- Difficulty level: Moderate
- Point of Error: Formula
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