Answer:
1 eV
Explanation:
Given:
Work function, ∅ = 2.00 eV
Kinetic energy of the ejected of the electron, K.E = 4.0 eV
Now,
using the photoelectric equation
, we have
Energy of the photon (E) = ∅ + K.E
also,
E = hc/λ
where, h is plank's constant
c is the speed of the light
λ is the wavelength
thus, we have
hc/λ = 2 + 4 = 6 eV
Energy of photon = 6eV
Now,
for the second case
λ' = 2λ
when Wavelength is doubled , E is halved
thus,
E' = hc/λ'
or
E' = hc/2λ
or
E' = E/2 = 6/2 = 3 eV
also,
E' = ∅ + KE
'
thus on substituting the values,
3 = 2 + KE'
or
KE' = 1 eV
Hence, the maximum kinetic energy for the second case is 1 eV
Answer:
The required solution is 100890 Pa and 14.3lb/in²
Explanation:
See attached image
Harmonics, Loop and Harmonic number
Hope this helps :)
Answer:
very low air pressure and that's not 100 points
