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:
Explanation:
mass m = 3 kg
spring constant be k
k x .8 = 40 N
k = 40 / .8 = 50 N /m
angular frequency ω = √ ( k / m )
= √ ( 50 / 3 )
= 4.08 rad /s
Let amplitude of oscillation be A .
1/2 k A² = 1/2 m v²
50 A² = 3 x 1²
A = .245 m = 24.5 cm
For displacement , the equation of SHM is
x = A sinωt
= 24.5 sin4.08 t
x = 24.5 sin4.08 t
Here, angle 4.08 t is in radians .
Answer:
13m/s. is the answer probably but do u have ms gallup too?