This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
Answer:

Explanation:
Let the linear charge density of the charged wire is given as

here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have

here we have


so we have

Answer:
It increases.
Explanation:
For the electron to escape the photon needs energy is equal to the difference between initial and its non quantised region energy , then only it will be able to escape finite well.
E ∝ n^2
n= energy state quantum number
so if , n increases maximum point of probability density increases.
To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.
The equation for the gravitational force between the Earth and the object on the surface of the Earth is

Where,
G = Universal gravitational constant
= Mass of Earth
= Distance between object and center of earth
= Mass of Object
The equation for the gravitational pulling force on the object due to gravitational acceleration is

Equation the two expression we have


This the acceleration due to gravity which is composite constant.
Replacing with our values we have then


The value of composite constant is
. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.