when a metal is irradiated by monochromatic light the maximum kinetic energy of the photoelectron is 1.2 electron volt if the frequency of the light is increased by 50% then kinetic energy of photoelectron is 3.6 electron volt evaluate the work function of metal
Answers
When metal is irradiated by monochromatic light the maximum kinetic energy of the photoelectrons is 1.2 electron volt. If frequency of light is increased 50% then maximum kinetic energy of the photoe.
full explanation.
⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ .
The maximum kinetic energy for the photoelectrons is
E
max
=hν−ϕ
where,ν is the frequency of incident light and ϕ is photoelectric work function of metal.
E
max
=h
λ
c
−ϕ .....................(since, ν=
λ
c
)
E
max
=
6600×10
−10
6.6×10
−34
×3×10
8
−1.5×1.6×10
−19
E
max
=3×10
−19
−2.4×10
−19
E
max
=0.6×10
−19
J
MY DEAR FRIEND PLEASE THANKS MY 15 ANSWERS ☺
When metal is irradiated by monochromatic light the maximum kinetic energy of the photoelectrons is 1.2 electron volt. If frequency of light is increased 50% then maximum kinetic energy of the photoe.
full explanation.
⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ .
The maximum kinetic energy for the photoelectrons is
E
max
=hν−ϕ
where,ν is the frequency of incident light and ϕ is photoelectric work function of metal.
E
max
=h
λ
c
−ϕ .....................(since, ν=
λ
c
)
E
max
=
6600×10
−10
6.6×10
−34
×3×10
8
−1.5×1.6×10
−19
E
max
=3×10
−19
−2.4×10
−19
E
max
=0.6×10
−19
J
The maximum kinetic energy for the photoelectrons is < /p > < p > Emax=hν−ϕ < /p > < p > where,ν is the frequency of incident light and ϕ is photoelectric work function of metal. < /p > < p > < /p > < p > Emax=hλc−ϕ .....................(since, ν=λc) < /p > < p > < /p > < p > Emax=6600×10−106.6×10−34×3×108−1.5×1.6×10−19 < /p > < p > < /p > < p > Emax=3×10−19−2.4×10−19 < /p > < p > < /p > < p > Emax=0.6×10−19J < /p > < p > < /p > < p > Themaximumkineticenergyforthephotoelectronsis </p><p>Emax=hν−ϕ</p><p>where,ν isthefrequencyof incidentlightand ϕ isphotoelectricworkfunctionofmetal.</p><p></p><p>Emax=hλc−ϕ .....................(since, ν=λc)</p><p></p><p>Emax=6600×10−106.6×10−34×3×108−1.5×1.6×10−19</p><p></p><p>Emax=3×10−19−2.4×10−19</p><p></p><p>Emax=0.6×10−19J</p><p></p><p>
MY DEAR FRIEND PLEASE THANKS MY 15 ANSWERS ☺
When metal is irradiated by monochromatic light the maximum kinetic energy of the photoelectrons is 1.2 electron volt. If frequency of light is increased 50% then maximum kinetic energy of the photoe.
full explanation.
⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ ⬇ .
The maximum kinetic energy for the photoelectrons is
E
max
=hν−ϕ
where,ν is the frequency of incident light and ϕ is photoelectric work function of metal.
E
max
=h
λ
c
−ϕ .....................(since, ν=
λ
c
)
E
max
=
6600×10
−10
6.6×10
−34
×3×10
8
−1.5×1.6×10
−19
E
max
=3×10
−19
−2.4×10
−19
E
max
=0.6×10
−19
J
The maximum kinetic energy for the photoelectrons is < /p > < p > Emax=hν−ϕ < /p > < p > where,ν is the frequency of incident light and ϕ is photoelectric work function of metal. < /p > < p > < /p > < p > Emax=hλc−ϕ .....................(since, ν=λc) < /p > < p > < /p > < p > Emax=6600×10−106.6×10−34×3×108−1.5×1.6×10−19 < /p > < p > < /p > < p > Emax=3×10−19−2.4×10−19 < /p > < p > < /p > < p > Emax=0.6×10−19J < /p > < p > < /p > < p > Themaximumkineticenergyforthephotoelectronsis </p><p>Emax=hν−ϕ</p><p>where,ν isthefrequencyof incidentlightand ϕ isphotoelectricworkfunctionofmetal.</p><p></p><p>Emax=hλc−ϕ .....................(since, ν=λc)</p><p></p><p>Emax=6600×10−106.6×10−34×3×108−1.5×1.6×10−19</p><p></p><p>Emax=3×10−19−2.4×10−19</p><p></p><p>Emax=0.6×10−19J</p><p></p><p>
MY DEAR FRIEND PLEASE THANKS MY 15 ANSWERS ☺
The actual question should be :
When a metal is irradiated by monochromatic light, the maximum kinetic energy of the photo-electrons is 1.2eV. If frequency of the light is increased 50% then maximum kinetic energy of photo-electron is 36eV Evaluate the work function of the metal.
Answer:
The required work function of the metal = 68.4 eV
Explanation:
We know that :
By Einstein Photo Electric Effect,
Km = hx - Ф
or, 1.2 = hυ - Ф
or, 1.2 + Ф = hυ
Now, it is given to us that the frequency of the light is increased by 50 %,
So, Km = hυ( 1 + ) - Ф = ( 3hυ / 2 ) - Ф
or, 36 = ( 3hυ / 2 ) - Ф
or, 36 = x ( 1.2 + Ф ) - Ф
or, Ф = 68.4 eV
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