Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.
The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:
E = 13.6 × Z² (1/n₂² - 1/n₁²) eV
where, n₁ is the initial energy level i.e. n₁ =7
n₂ is the higher energy level i.e. n₂ = 4
E is the energy possessed
Z is the atomic number, Z = 1 for H-atom
Subsituting in above equation,
E = 13.6 (1/16 - 1/49) eV
E = 0.27 eV
We know that,
E = hc / λ
where, h is Planck constant
c is speed of light
λ is wavelength
On subsituting,
0.27 eV = 1240/ λ
⇒ λ = 4592.59 nm
Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.
Learn more about Energy of Photon here, brainly.com/question/2393994
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Answer:
1.7 seconds
Explanation:
To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m
first we need to find the initial speed to just enter the intersection by using the third equation of motion
v^2 - u^2 = 2*a*s
45^2 - u^2 = 2 * -5.7 * 84.7
u^2 = 45^2 +965.58
u^2 = 2990.58
u = 54.7 m/s
Now for time we apply the first equation of motion
v-u =a * t
t = (v-u)/a = (45 - 54.7)/-5.7 = 1.7seconds
Similarities:
The halogens like noble gases are gaseous im nature. Example,
Chlorine, bromine are halogens and argon , xenon are noble gases.
Both are non-metals.
Differences: Halogens are very reactive because their octet needs only one electron tk complete. But, the nobles gases are quite stable and unreactive because the have complete octets.
A right turn would be the answer probably.
