Atomic Structure(11th And 12th > Chemistry ) Questions and Answers
Explanation:-
Answer: Option C. -> 2 × 10−29 kg:
C
Δx × m × Δv ≥ h4π
1 × 10−10 × m × 3.34π × 105 ≥ 6.6 × 10−344 × π
M = 2 × 10−29 kg
When a certain metal was irradiated with light of frequency 4.0 × 1016 s−1, the photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradicated with light of frequency 2.0 × 1016s−1. Calculate the critical frequency (ν0) of the metal.
Explanation:-
Answer: Option D. -> 1 × 1016 s−1:
D
We know that, the incident photons should have a minimum frequency called threshold frequency and a certain amount of energy required called work function. Therefore,
KE = hν − hν0 = h(ν − ν0)
KE1 = h(ν1 − ν0) ...(i)
KE2 = h(ν2 − ν0) ...(ii)
Dividing equation (ii) by (i), we get
KE2KE1 = h(ν2 − ν0)h(ν1 − ν0) = (ν2 − ν0)(ν1 − ν0)
But given that
KE2KE1 = 3
∴ 3 = (ν2 − ν0)(ν1 − ν0) ⇒ 3(ν1 − ν0) = ν2 − ν0
⇒ 3ν1 − ν2 = 3ν0 − ν0 = 2ν0
⇒ 3 × 2.0 × 1016 − 4 × 1016 = 2ν0
⇒ ν0 = 2 × 10162 = 1 × 1016s−1
Explanation:-
Answer: Option D. -> Will be reduced by 25%:
D
No change by doubling mass of electrons however by reducing mass of neutron to half total atomic mass becomes 6+3 instead of 6+6. Thus reduced by 25%.