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vovikov84 [41]
4 years ago
9

The energy required to dislodge electrons from sodium metal via the photoelectric effect is 275 kj/mol. What wavelength in nm of

light has sufficient energy per photon to dislodge an electron from the surface of sodium?
Physics
1 answer:
denis23 [38]4 years ago
7 0

Answer: 430 nm

Explanation:

For 257 kJ to dislodge one mole of electrons we need,

275 x 10^3 / 6 x 10^23 = 4.6 x 10^-19

Using Einstein’s relationship between energy, frequency and wavelength

E = hf = h x (c/landa)

Therefore

Landa = h x (c/E) = 6.6 x 10^-34 x (3 x 10^8 / 4.6 x 10^-19 = 4.3 x 10-7 m.

In nm, landa = 430 nm

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Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
The express train can travel 100 miles per hour. This information describes the trains
Andreas93 [3]
Rate of speed, probs is the answer
5 0
3 years ago
Please refer to the picture
Vinvika [58]
I think is c not sure though
3 0
3 years ago
A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
3 years ago
Famed stunt pilot, Cleonvia Thread, pulls out rapidly from a dive. He is traveling at 222 mi/h at the bottom of his trajectory,
Simora [160]

Answer:

0.39 m/s²

Explanation:

From the question,

a = v²/r.................... Equation 1

Where v = velocity, r = radius.

Given: v = 222 mi/h = (222×0.44704) m/s = 9.83 m/s, r = 820 ft = (820×0.3048) m = 249.94 m.

Substitute thses values into equation 1

a = 9.83²/249.94

a = 96.63/249.94

a = 0.39 m/s²

Hence the  acceleration is 0.39 m/s²

8 0
3 years ago
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