A magnesium surface has a work function of 3.60 eV. Electromagnetic waves with a wavelength of 320 nm strike the surface and eje
1 answer:
Answer:
K(maximum) = 0.284 eV
Explanation:
Given data
work function ω = 3.60 eV = 3.60 × 1.6 × = 5.76 ×
J
wavelength λ = 320 nm = 320 × m
to find out
maximum kinetic energy
solution
we know that speed of light is = 3 × m/s
and planck constant h = 6.63 × J.s
we will apply here Einstein's photo elecric equation that is
hc/λ = ω + K(maximum)
put here all these value we get
6.63 × × 3 ×
/ 320 ×
= 5.76 ×
+ K(maximum)
so
K(maximum) = 6.63 × × 3 ×
/ 320 ×
- 5.76 ×
K(maximum) = 0.4556 × J
K(maximum) = 0.4556 × / 1.6 ×
K(maximum) = 0.284 eV
You might be interested in
Answer:
A. when the mass has a displacement of zero
Explanation:
The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:
where
k is the spring constant
x is the displacement of the mass with respect to the equilibrium position of the spring
m is the mass
v is the velocity of the mass
Since the total energy E must remain constant, we can notice the following:
- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum
- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0
Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.