PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<
PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<
1st Edition
ISBN: 9781323834831
Author: Knight
Publisher: PEARSON C
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Chapter 38, Problem 1CQ

(a).

To determine

Thedischarge of charged electroscope by shining in ultraviolet light

(a).

Expert Solution
Check Mark

Answer to Problem 1CQ

Solution: The discharge of charged electroscope by shining in ultraviolet light is due to Photoelectric effect.

Explanation of Solution

Given Info: Negatively charged electroscope
Ultraviolet light

The discharge is due to Photoelectric effect. According to Einstein’s photoelectric equation, each photon in incident light will bring an electron on top of metal surface (which is called as work function) and the leftover energy of light is given to photo electrons as kinetic energy. In the present case, incident light energy of ultraviolet should be greater than work function of negatively charged electroscope and this causes discharge.

Conclusion:

Thedischarge of charged electroscope by shining in ultraviolet lightis due to Photoelectric effect.

(b).

To determine

Thedischarge of uncharged electroscope by shining in ultraviolet light may not make positively charged electroscope.

(b).

Expert Solution
Check Mark

Answer to Problem 1CQ

Solution:

In the present case of uncharged electroscope, incident light energy of ultraviolet should be lesser than work function and this cause non -discharge of electroscope. Hence, the electroscope may not become positive.

Explanation of Solution

Given Info: Uncharged electroscope
Ultraviolet light

In the present case of uncharged electroscope, incident light energy of ultraviolet should be lesser than work function and this cause non discharge of electroscope. Hence, the electroscope may not become positive.

Conclusion:
In the present case of uncharged electroscope, incident light energy of ultraviolet should be lesser than work function and this cause non discharge of electroscope. Hence, the electroscope may not become positive.

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Taking a Hike A hiker begins a trip by first walking 21.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 46.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger's tower. y (km) Can N W-DE 45.0° 60.0° Tent Tower B x (km) ☹ (a) Determine the components of the hiker's displacement for each day. SOLUTION Conceptualize We conceptualize the problem by drawing a sketch as in the figure. If we denote the displacement vectors on the first and second days by A and B, respectively, and use the ---Select-- as the origin of coordinates, we obtain the vectors shown in the figure. The sketch allows us to estimate the resultant vector as shown. Categorize Drawing the resultant R, we can now categorize this problem as one we've solved before: --Select-- of two vectors. You should now have a hint of the power of categorization in that many new problems are very similar to problems we have already solved if we are…

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PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<

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