Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.

Answers

Answer 1
Answer:

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

E = (13.6Z^(2) )/(n^(2) ) eV     .....(1)

Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for Z and 1 for n in the equation (1).

E = (13.6*(6)^(2) )/(1^(2) )

E = 489.6 eV

The wavelength (λ)  of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

(1)/(\lambda) =RZ^(2)[(1)/(n_(1) ^(2))-(1)/(n_(2) ^(2) )]     ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

(1)/(\lambda) =1.09*10^(7) *6^(2)[(1)/(2 ^(2))-(1)/(3 ^(2) )]

(1)/(\lambda)  = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m


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An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pulls his arms towards his body to reduce his MOMENT OF INTERTIA by 1/2, what is his ANGULAR VELOCITY w? c) How much does his ROTATIONAL KINETIC ENERGY change? That is, if the initial Kinetic Energy is (KE)initial, what is the final KE? d) Where did that ENERGY come from, or go to?

Answers

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final =  1/2 I₂ ω₂²

= 1/2 (I₁ / 2)  (2ω)²

=  I₁ ω²

c )

Change in rotational kinetic energy

=  I₁ ω² -  1/2 I₁ ω²

=  +  1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF when connected in series. What is the capacitance of each capacitor?

Answers

Answer:

C_1=7.23pF\ and\ C_2=2.19pF

Explanation:

Let the two capacitance are C_1\ and\ C_2

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So C_1+ C_2=9.2--------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So (1)/(C_1)+(1)/(C_2)=(1)/(1.68)

(C_1+C_2)/(C_1C_2)=(1)/(1.68)

C_1C_2=1.68* 9.42=15.8256pF

C_1-C_2=√((C_1+C_2)^2-4C_1C_2)=√(9.42^2-4* 15.8256)=5.0432pF-----EQN

On solving eqn 1 and eqn 2

C_1=7.23pF\ and\ C_2=2.19pF

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

   v = 4.1 m / s

Explanation:

Velocity is defined by the relation

          v =(dx)/(dt)

 we perform the derivative

         v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

        v = (\Delta x)/(\Delta t)

        Δx = v Δdt + x₀

        h=   4.1 t + 5.5

       

         v = 4.1 m / s

         x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.

Answers

The anchoring force needed to hold the elbow in place is 839.5 N.

How to find anchoring force?

To determine the anchoring force needed to hold the elbow in place, use the following steps:

Apply the conservation of momentum equation to the elbow:

∑F = ˙m(V₂ - V₁)

where:

˙m = mass flow rate

V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively

F = anchoring force

The mass flow rate is given by:

˙m = ρAV

where:

ρ = density of water (1000 kg/m³)

A = cross-sectional area of the elbow

V = velocity

The velocities at the inlet and outlet of the elbow can be calculated using the following equations:

V₁ = A₁V₁

V₂ = A₂V₂

where:

A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively

Calculate the momentum flux correction factor at the inlet and outlet of the elbow:

β₁ = 1.03

β₂ = 1.03

Substitute all of the above equations into the conservation of momentum equation:

F = ˙m(V₂ - V₁)

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

Calculate the velocity at the inlet of the elbow:

V₁ = A₂V₂/A₁

V₁ = (25 cm²/150 cm²)(V₂)

V₁ = 1/6 V₂

Substitute the above equation into the conservation of momentum equation:

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)

F = ρV₂²(1.03)(1/36 A₁ - A₂)

Calculate the anchoring force:

F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)

F = 839.5 N

Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.

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The anchoring force needed to hold the elbow in place is 932 N.

The anchoring force needed to hold the elbow in place is the sum of the following forces:

The force due to the change in momentum of the water as it flows through the elbow.

The force due to the weight of the elbow and the water in it.

The force due to the buoyancy of the water in the elbow.

The force due to the change in momentum of the water can be calculated using the momentum equation:

F = mΔv

where:

F is the force

m is the mass of the fluid

Δv is the change in velocity of the fluid

In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:

m = ρAv

where:

ρ is the density of the fluid

A is the cross-sectional area of the pipe

v is the velocity of the fluid

The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

where:

P1 is the pressure at the inlet

v1 is the velocity at the inlet

P2 is the pressure at the outlet

v2 is the velocity at the outlet

In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:

A1v1 = A2v2

where:

A1 is the cross-sectional area at the inlet

A2 is the cross-sectional area at the outlet

The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:

W = mg

where:

W is the weight

m is the mass

g is the acceleration due to gravity

The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:

B = ρVg

where:

B is the buoyancy

ρ is the density of the fluid

V is the volume of the fluid displaced

g is the acceleration due to gravity

The volume of the fluid displaced by the elbow is equal to the volume of the elbow.

Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.

F_anchor = F_momentum + F_weight - F_buoyancy

where:

F_anchor is the anchoring force

F_momentum is the force due to the change in momentum of the water

F_weight is the force due to the weight of the elbow and the water in it

F_buoyancy is the force due to the buoyancy of the water in the elbow

Plugging in the values for each force, we get:

F_anchor = 1030 N - 490 N + 392 N = 932 N

Therefore, the anchoring force needed to hold the elbow in place is 932 N.

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Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answers

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

K_(max)=(hc)/(\lambda)-W

Here h is the Planck's constant, c is the speed of light, \lambda is the wavelength of the light and W the work function of the element:

W=(hc)/(\lambda)-K_(max)\nW=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\nW=2.01eV

Now, we calculate the wavelength for the new maximum kinetic energy:

W+K_(max)=(hc)/(\lambda)\n\lambda=(hc)/(W+K_(max))\n\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\n\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}

The momentum (P₂) after a collision will be

\rm P_2 = (m_1 + m_2) u \ cos\  \theta \  \hat{x} \ + (m_1 + m_2) u \ sin \  \theta \  \hat{y}

Applying momentum conservation, we have

\rm  (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\  \theta \   \n\n  ...1

\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \  \theta  ...2

From equations 1 and 2, we have

\rm \theta =  tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta =  tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o

From equation 1, we have

\rm u =    ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\  \theta )  \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s

Then the change in kinetic energy, we have

\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J

The fraction of kinetic energy lost will be

\rm Energy \ lost = (239.34)/(267.5) = 0.895

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Answer:

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