Physics, asked by anunyathatikonda, 6 months ago

8.23: The minimum energy required to overcome
the attractive forces between an electron and
the surface of Ag metal is 5.52 x 10-19 J. What
will be the maximum kinetic energy of
electron ejected out from Ag which is being
exposed to UV-light of 2 = 360 Aº ?​

Answers

Answered by Anonymous
0

Answer:

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Sample testcasesJumping

numbers but 796 and 89098 are notin sorted

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Sample testcasesJumping

numbers but 796 and 89098 are notin sorted

Sample testcasesJumpingattractive forces between an electron and

the surface of Ag metal is 5.52 x 10-19 J. What

will be the maximum kinetic energy of

electron ejected out from Ag which is being

exposed to UV-light of 2 = 360 Aº ?

numbers but 796 and 89098

Answered by rohankaurav05
2

Answer:

K.E = hc/lamda -work function

given : work function=5.52 ×10^-19

K.E = [(6.626×10^-34 ×3×10^8) /360 10^-10]-5.52×10^-19

= 49.68 × 10^-19

K.E= 49.68 ×10^-19

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