Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.
Determine the friction coefficient.
Multiply these values by each other: 250 N * 0.13 = 32.5 N .
You just found the force of friction!
Explanation:
Answer: A 120 metros por segundo
Explanation: multiplicas la velocidad por el tiempo
Answer:
option C
Explanation:
given,
Force on the object = 10 N
distance of push = 5 m
Work done = ?
we know,
work done is equal to Force into displacement.
W = F . s
W = 10 x 5
W = 50 J
Work done by the object when 10 N force is applied is equal to 50 J
Hence, the correct answer is option C