stA student investigated the photoelectric effect and found that the threshold frequency for atheparticular metal was 6.0x1014 Hz. When light with a frequency of 9.0 x 1 0 Hz was shone onto theThe student was surprised to findlesame metal surface the stopping voltage was recorded as 0.42that doubling the frequency of the incident radiation did not double the stopping voltage, nor did itdouble the current flowing.leAccount for the student's observations.1Srec9 MVe

Question
Asked Oct 31, 2019

Question in image. Disregard the pen markings plz.

st
A student investigated the photoelectric effect and found that the threshold frequency for a
the
particular metal was 6.0x1014 Hz. When light with a frequency of 9.0 x 1 0 Hz was shone onto the
The student was surprised to find
le
same metal surface the stopping voltage was recorded as 0.42
that doubling the frequency of the incident radiation did not double the stopping voltage, nor did it
double the current flowing.
le
Account for the student's observations.
1
Srec
9 M
Ve
help_outline

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st A student investigated the photoelectric effect and found that the threshold frequency for a the particular metal was 6.0x1014 Hz. When light with a frequency of 9.0 x 1 0 Hz was shone onto the The student was surprised to find le same metal surface the stopping voltage was recorded as 0.42 that doubling the frequency of the incident radiation did not double the stopping voltage, nor did it double the current flowing. le Account for the student's observations. 1 Srec 9 M Ve

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check_circleExpert Solution
Step 1

Write the expression for stopping potential

(f-f
V
e
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(f-f V e

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Step 2

Substitute the values to find f

(6.626x10 J/sf-(6.0x 10 Hz)
0.42 V
1.602x10 C
(1.602x10-19 C) (0.42 V)
(6.626x10 J/s)
f = (6.0x10 Hz)+
= 7.0 x1014 Hz
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(6.626x10 J/sf-(6.0x 10 Hz) 0.42 V 1.602x10 C (1.602x10-19 C) (0.42 V) (6.626x10 J/s) f = (6.0x10 Hz)+ = 7.0 x1014 Hz

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Step 3

Write the expression for new stopping pot...

h(2f-fo
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h(2f-fo

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