Question

Question:4 ) 

Consider the following results, obtained using 5.0 eV photons and the same photocell that is discussed in the passage.

Relative
intensity
of light

Electrical
current (mA)

Maximum kinetic
energy of ejected
electron (eV)

low. 28 3.1

medium. 42 3.1

high. 58 3.1



The maximum kinetic energy of the ejected electron, 3.1 eV, was not the expected value. The expected value was:

A)    0.0 eV
B)    between 0.1 eV and 0.8 eV
C)    between 0.9 eV and 2.9 eV
D)    greater than 3.0 eV




please tell me quickly
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Answer 1

Answer:

option C is the right answer

Explanation:

Kinetic Energy lost due to stopping potential =e×3.5 V

=3.5eV

So,

From the question, the photoelectric equation becomes,

5−W=3.5

⇒W=1.5 eV

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