calculate the stopping potential for photoelectrons which is emitted when 400 nm wavelength of photon falls on metal surface whose threshold wavelength is 600 nm
= 19.878 x 10^-26 J·m * [(1 / 400 - 1 / 600) x 10^9 m^-1]
= 19.878 x 10^-17 * (3/12) x 10^9 V
= 1.316 x 10^-8 V
Therefore, the stopping potential for photoelectrons emitted when a 400 nm wavelength photon falls on a metal surface with a threshold wavelength of 600 nm is approximately 1.316 x 10^-8 V.
>>Stopping Potential and Einstein's Photoelectric Equation
>>A light of wavelength 600 nm is incident
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A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metal is:
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Explanation:
The stopping potential for photoelectrons can be calculated using the equation:
Stopping Potential (V) = (hc / λ) - (hc / λ₀)
Where:
- V is the stopping potential,
- h is Planck's constant (6.626 x 10^-34 J·s),
- c is the speed of light (3.0 x 10^8 m/s),
- λ is the wavelength of the incident photon,
- λ₀ is the threshold wavelength of the metal surface.
Given:
- λ = 400 nm = 400 x 10^-9 m
- λ₀ = 600 nm = 600 x 10^-9 m
Let's calculate the stopping potential using the given values:
V = (hc / λ) - (hc / λ₀)
= [(6.626 x 10^-34 J·s) * (3.0 x 10^8 m/s)] / (400 x 10^-9 m) - [(6.626 x 10^-34 J·s) * (3.0 x 10^8 m/s)] / (600 x 10^-9 m)
Simplifying the equation:
V = [(6.626 x 10^-34 J·s) * (3.0 x 10^8 m/s)] * [(1 / 400 x 10^-9 m) - (1 / 600 x 10^-9 m)]
= (6.626 x 3.0) * [(10^-34 J·s) * (10^8 m/s)] * [(1 / 400 - 1 / 600) x 10^9 m^-1]
= 19.878 x 10^-26 J·m * [(1 / 400 - 1 / 600) x 10^9 m^-1]
= 19.878 x 10^-17 * (3/12) x 10^9 V
= 1.316 x 10^-8 V
Therefore, the stopping potential for photoelectrons emitted when a 400 nm wavelength photon falls on a metal surface with a threshold wavelength of 600 nm is approximately 1.316 x 10^-8 V.
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Class 12
>>Physics
>>Dual Nature of Radiation and Matter
>>Stopping Potential and Einstein's Photoelectric Equation
>>A light of wavelength 600 nm is incident
Question
Bookmark
A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metal is:
Hard
Updated on : 2022-09-05
Solution
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Correct option is A)
Here, λ=600nm,λ
′
=400nm,K
max
′
=2K
max
According to Einstein photoelectric equation
K
max
=
λ
hc
−ϕ
0
and 2K
max
=
λ
′
hc
−ϕ
0
Dividing by we get
2=
λ
hc
−ϕ
0
λ
′
hc
−ϕ
0
or
λ
2hc
−2ϕ
0
=
λ
′
hc
−ϕ
0
λ
2hc
−2ϕ
0
=
λ
′
hc
−ϕ
0
Or hc(
λ
2
−
λ
′
1
)=ϕ
0
∴ϕ
0
=1240eVnm(
600nm
2
−
400nm
1
)=1.03eV