Answer:
Frequency, 
Explanation:
Given that,
The wavelength of the x-rays, 
We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :
f is the frequency
So, the frequency of an x-ray is 
. Hence, this is the required solution.
Answer: A: electron shells outside a central nucleus
B: hard, indivisible sphere
C: mostly empty space
Which list of atomic model descriptions represents
the order of historical development from the earliest
to most recent?
Explanation:
3
Answer:
a)The electric Field will be zero at the point between the sheets
b)
c)
Explanation:
Let 
be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.
Now using Gauss law we have, E be the electric Field at the distance r from the sheet then
The Field will be away from the sheet and perpendicular to it.
a) The Electric Field between them
b)The Electric Field to the right of the sheets
c)The Electric Field to the left of the sheets
Answer:
True
Explanation:
If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell
According to Gauss law
∅ = EA =-Q/∈₀
Where ∅ is the electric flux through the gaussian surface and E is the electric field strength
If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero
Becomes a +1 ion for this
