Name: Planck's Quantum Theory
Uploaded: 2019-07-19
Duration: 4 min 5 s
Description: Planck made many contributions to theoretical physics, but his fame rests primarily on his role as originator of the quantum theory. Watch this video to lear...

Question

When the light of

3000 \overset{o}{A}

is incident on sodium cathode, the stopping potential is

1.85 V

; and when the light of

4000 \overset{o}{A}

is incident, then the stopping potential becomes

0.82 V

. Calculate Planck's constant, work function of sodium, threshold wavelength for sodium.

Accepted Answer

The photoelectric equation can be written as:

$hν = K_{\max} + ϕ$ where $h, ν$ are the Planck's constant and the frequency of incident radiations and $ϕ$ = work function of the metal. This can also be written as for two wave lengths as:

$\frac{hc}{λ_{1}} = ϕ + {eV}_{1} and \frac{hc}{λ_{2}} = ϕ + {eV}_{2}$ where $V_{1 .} V_{2}$ are the stopping potential for wavelength $λ_{1} a n d λ_{2}$ we get by subtracting

$h c [\frac{1}{λ_{2}} - \frac{1}{λ_{1}}] = e [V_{2} - V_{1}]$

Given here,

$c = 3 \times 10^{8} m / s, λ_{1} = 3 \times 10^{- 7} m and λ_{2} = 4 \times 10^{- 7} m$

$V_{1} = 0, 82 V V_{2} = 1.85 V$

Substituting the values,

$h \times 3.0 \times 10^{8} [\frac{1}{3 \times 10^{- 7}} - \frac{1}{4 \times 10^{- 7}}] = 1.6 \times 10^{- 19} (1.85 - 0.82)$

or $h \times 3.0 \times 10 \times \frac{1}{12 \times 10^{- 7}} = 1.6 \times 10^{- 19} (1.03)$

or $h = \frac{1.6 \times 10^{- 19} \times 1.03 \times 12 \times 10^{- 7}}{3 \times 10^{8}} = 6.6 \times 10^{- 34} Js$

The maximum kinetic energy in the first case is given by: $K_{\max} = \frac{hc}{λ} - ϕ = \frac{6.62 \times 10^{- 34} \times 3.0 \times 10^{8}}{3 \times 10^{- 7} \times 1.6 \times 10^{- 19}} - 1.85 = 4.13 - 1.85 = 2.28 eV$

The stopping potential $(e V_{0})$ will completely stop the electron with this energy so we have ${eV}_{0} = 2.28 eV or v_{0} = 2.28 V$

The Stopping potential will be the light of wavelength corresponding to this potential i.e $\frac{hc}{λ_{0}} = 2.28 \times e or λ_{0} = \frac{hc}{e \times 2.28}$

Substituting the values, we have

$λ_{0} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.28 \times 1.6 \times 10^{- 19}} = 5440 \times 10^{- 10} m$

When the light of $3000 \overset{o}{A}$ is incident on sodium cathode, the stopping potential is $1.85 V$ ; and when the light of $4000 \overset{o}{A}$ is incident, then the stopping potential becomes $0.82 V$ . Calculate Planck's constant, work function of sodium, threshold wavelength for sodium.

Important Questions on Photoelectric Effect

Learn and Prepare for any exam you want !

When the light of 3000 Ao is incident on sodium cathode, the stopping potential is 1.85 V; and when the light of 4000 Ao is incident, then the stopping potential becomes 0.82 V. Calculate Planck's constant, work function of sodium, threshold wavelength for sodium.

Important Questions on Photoelectric Effect

Learn and Prepare for any exam you want !

When the light of $3000 \overset{o}{A}$ is incident on sodium cathode, the stopping potential is $1.85 V$ ; and when the light of $4000 \overset{o}{A}$ is incident, then the stopping potential becomes $0.82 V$ . Calculate Planck's constant, work function of sodium, threshold wavelength for sodium.