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Lady_Fox [76]
3 years ago
8

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.4×1015hz?

Chemistry
1 answer:
Anika [276]3 years ago
7 0
<span>The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.</span>
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When an electron passes through the magnetic field of a horseshoe magnet, the electron's?
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When an electron passes through the magnetic field of a horseshoe magnet, the electron's direction is changed.

Path of an electron in a magnetic field

The force (F) on wire of length L carrying a current I in a magnetic field of strength B is given by the equation:

F = BIL

But Q = It and since Q = e for an electron and v = L/t you can show that :

Magnetic force on an electron = BIL = B[e/t][vt] = Bev where v is the electron velocity

In a magnetic field the force is always at right angles to the motion of the electron (Fleming's left hand rule) and so the resulting path of the electron is circular.

Therefore :

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8 0
3 years ago
Read 2 more answers
Help help help !!!!!!
11111nata11111 [884]

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V

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