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Novay_Z [31]
2 years ago
14

Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with

Z=5 and only a single electron orbiting the nucleus. This expression predicts equal electron energies for these two species for certain values of the quantum number n (the quantum number is different for each species). For quantum numbers less than or equal to 9, what is the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level?
Physics
1 answer:
ella [17]2 years ago
3 0

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

so

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= -  13.6 x2² / 8²

= -  0.85 eV .

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Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

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h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
If a body is moving in the horizontal axis with a velocity Vx= 6m/s and in the vertical axis Vy=8m/s What is the angle Theta abo
cluponka [151]

Answer: C

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2 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

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Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
2 years ago
A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

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where

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\Delta t = 2.0 ms = 0.002 s is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

\Phi_B = BA

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2

At the beginning, the magnetic field is

B_i = 1.5 T

so the flux is

\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

\Phi_f = 0

so the variation of magnetic flux is

\Delta \Phi = 0-0.068 Wb=-0.068 Wb

And substituting into (1) we find the average emf in the coil

\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0
2 years ago
A piece of copper of mass 100 g is being drilled through with a 1/2" electric drill. The drill operates at 40.0 W and takes 30.0
lesantik [10]

Answer:

correct option is c. 31.0°C  

Explanation:

given data

copper of mass = 100 g

electric drill = 1/2"

power = 40.0 W

time = 30 s

C copper = 387 J/kgC

to find out

copper's increase in temperature

solution

we get here energy that is express a s

energy = 40 W × 30 s

energy = 1200 Watt seconds

and heat acquired by drill is here as

heat acquired = 100 × T × 387  

here temperature rise in copper mass as

temperature rise in copper mass = \frac{100}{1000} × T × 387

temperature rise in copper mass = 38.7 ×  T Watt second

we know that all the energy from the drill heats the copper

so we can say

38.7 ×  T = 1200

T = 31°C  

so correct option is c. 31.0°C  

4 0
2 years ago
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