Question
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.
Answer: Option D
:
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
Was this answer helpful ?
:
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
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution