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podryga [215]
4 years ago
11

Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the bind

ing energy of electrons to sodium metal?
Physics
1 answer:
Travka [436]4 years ago
4 0

Answer:

Binding Energy = 2.24 eV

Explanation:

First, we need to find the energy of the photon of light:

E = hc/λ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)

E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 3.1 eV

Now, from Einstein's Photoelectric Equation:

E = Binding Energy + Kinetic Energy

Binding Energy = E - Kinetic Energy

Binding Energy = 3.1 eV - 0.86 eV

<u>Binding Energy = 2.24 eV</u>

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Some substances look alike and the density can help you determine which is which. It also helps to tell which is heavier than the other.
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4 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol
bearhunter [10]

Answer:

 N= 3

Explanation:

For this exercise we must use Faraday's law

          E = - dФ / dt

         Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

         E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

         cos 0 = 1

         A = π r²

   

In the length of the wire there are N turns each with a length L₀ = 2π r

          L = N (2π r)

          r = L / 2π N

    we substitute

          A = L² / (4π N²)

The magnetic field produced by a solenoid is

           B = μ₀ N/L   I

for which

            B₀ = μ₀  N/L   I

           

The final field is zero, because the current is zero

            B = 0

We substitute

           E = - (L² / 4π N²)  (0 - μ₀ N/L I) / t

           E = μ₀ L I / (4π N t)

           N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

           N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

           N  = 320 10⁻⁷ / 9.6 10⁻⁶

           N = 33.3 10⁻¹

          N= 3

           

7 0
3 years ago
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Ilya [14]

Elliptical orbit.<<<<<<<<<<



8 0
3 years ago
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

x = v_x t

97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

7 0
4 years ago
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