Each principal energy level has one sublevel containing one orbital, an s orbital, that can contain a maximum of two electrons. Electrons in this orbital are called s electrons and have the lowest energy of any electrons in that principal energy level.
Answer:
Explanation:
Initially no of atoms of A = N₀(A)
Initially no of atoms of B = N₀(B)
5 X N₀(A) = N₀(B)
N = N₀
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀
For B
N(B) = N(B)₀
N(A) = N(B) , for t = 2 h
N(A)₀ = N(B)₀
N(A)₀ = 5 x N₀(A)
= 5
= 5
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
= 5
Putting t = 2 h , λ₁ = .9 h⁻¹
= 5
= 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .
Heya!!!
Answer to your question:
All you need to do is to compare both the equations...
On comparing, you'll get
d=6m
v=6.5m/s
a=-9.81m/s^2
we know,
6= 6.5t -0.5 (9.81)t^2
6=6.5t-4.905 t^2
4.905 t^2-6.5t +6=0
now you can use quadratic eq. to solve this.
Hope it helps ^_^
I just figured this out now.
First you would use the formula
Ephoton= hc/λ and substitute in the value's of plank's constant, the speed of light in a vaccum and the wavelength which will give you the energy in joules. Then you go to the reference table and solve for the energy used between the different levels for Mercury making sure to convert electron volts to jules. In the end the correct answer should be energy level D.
To solve the problem, start by applying the concepts related to current in an RL circuit. The current is defined exponentially and using Ohm's law we can put the initial current in terms of the voltage and resistance. Consecutively with the calculated time constant we can find the respective inductance. For the second part we will apply the electrical potential energy connectors to find the amount of stored energy.
PART A)
Inductance can be defined then,
PART B) Now the energy is given under the terms:
Therefore the energy stored in the coil at this same moment is 0.0002727J