If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of l

Q: Why did we completely ignore the work function $\phi$?

The problem explicitly states that the kinetic energy is "very high in comparison to work function". In physics, when a quantity is significantly smaller than another in an addition or subtraction ($KE \gg \phi$), it can be approximated to zero to simplify calculations without losing much accuracy.

Q: How did we get $KE = p^2/2m$?

Kinetic energy is $KE = \frac{1}{2}mv^2$. Momentum is $p = mv$. If you square the momentum, you get $p^2 = m^2v^2$. Dividing this by $2m$ gives $\frac{m^2v^2}{2m} = \frac{1}{2}mv^2$, which is exactly the kinetic energy.

Q: Why does the question specify the "fastest" electron?

Electrons inside a metal are bound by varying amounts of energy. The "fastest" electron is the one that was bound by the minimum possible energy, which is exactly the work function $\phi$. Einstein's equation $E = KE_{max} + \phi$ specifically applies to these fastest electrons.

Q: Is the wavelength $\lambda$ here the de Broglie wavelength of the electron?

No, be careful! The $\lambda$ in the question refers to the wavelength of the *incident light* (photon), not the de Broglie wavelength of the ejected electron. The de Broglie wavelength of the electron would be $\lambda_e = h/p$.

Q: What if the momentum was doubled instead of becoming $1.5$ times?

Since $\lambda \propto 1/p^2$, if the momentum is doubled ($2p$), the new wavelength would become $1/2^2 = 1/4$ of the original wavelength. The inverse square relationship means the wavelength decreases rapidly as momentum increases.

Question

If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

Options:

(A) $\frac{4}{9} \lambda$

(B) $\frac{3}{4} \lambda$

(C) $\frac{2}{3} \lambda$

(D) $\frac{1}{2} \lambda$

Sigma Percentile · Accepted Answer

### The Photoelectric Setup

Imagine a photon of light acting like a tiny billiard ball, striking a metal surface and knocking an electron out. This is the essence of the **photoelectric effect**.

The energy of the incoming photon is split into two parts: overcoming the metal's binding energy (the **work function**, $\phi$), and giving the ejected electron its kinetic energy ($KE$).

Mathematically, this is Einstein's famous equation:

$$E = KE + \phi$$

### The Crucial Approximation

The problem gives us a massive shortcut. It states that the kinetic energy of the ejected photoelectron is *very high* in comparison to the work function.

What does this mean for our equation? It means $\phi$ is negligibly small compared to $KE$.

We can safely approximate the equation to:

$$E \approx KE$$

This simplifies our life immensely!

### Bridging Energy and Momentum

Now, we need to connect the properties of the light to the properties of the electron.

For the incident light, the energy of a photon is given by $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

For the ejected electron, its kinetic energy can be expressed in terms of its momentum $p$. Using the relation $KE = \frac{1}{2}mv^2$ and $p = mv$, we get $KE = \frac{p^2}{2m}$.

### The Master Equation

Let's bring it all together. By equating the photon's energy to the electron's kinetic energy, we get:

$$ \frac{hc}{\lambda} = \frac{p^2}{2m} $$

Rearranging this to solve for wavelength $\lambda$, we find:

$$ \lambda = \frac{2mhc}{p^2} $$

Notice that $2$, $m$, $h$, and $c$ are all constants. This reveals a beautiful inverse square relationship:

**$\lambda \propto \frac{1}{p^2}$**

### Final Calculation

We are asked to find the new wavelength $\lambda_2$ when the momentum becomes $1.5p$.

Using our proportionality, we can set up a ratio between the two states:

$$ \frac{\lambda_2}{\lambda_1} = \left(\frac{p_1}{p_2}\right)^2 $$

Substitute the given values: $p_1 = p$ and $p_2 = 1.5p$.

$$ \frac{\lambda_2}{\lambda} = \left(\frac{p}{1.5p}\right)^2 $$

Since $1.5 = \frac{3}{2}$, the ratio becomes $\frac{2}{3}$. Squaring this gives us:

$$ \frac{\lambda_2}{\lambda} = \frac{4}{9} $$

Therefore, the new wavelength is **$\frac{4}{9} \lambda$**. This perfectly matches option (a).

Select Answer:

Visualized Solution

$P h o t oe l ec t r i c S e t u p$

$A pp r o x ima t i o n$

$E n er g y an d M o m e n t u m$

$E q u a t in g T er m s$

$P r o p or t i o na l i t y$

$R a t i o S e t u p$

$S u b s t i t u t i o n$

$F ina l R es u l t$

$F oo d f or T h o ug h t$

The Sigma Insight: Wave Particle Duality

The Photoelectric Setup

The Crucial Approximation

Bridging Energy and Momentum

The Master Equation

Final Calculation

Similar Questions

The de-Broglie wavelength ( $λ$ ) associated with a photoelectron varies with the frequency ( $ν$ ) of the incident radiation as, [ $ν_{0}$ is threshold frequency]

The work function of sodium metal is $4.41 \times 1 0^{- 19}$ J. If photons of wavelength $300$ nm are incident on the metal, the kinetic energy of the ejected electrons will be $(h = 6.63 \times 1 0^{- 34} J s; c = 3 \times 1 0^{8} m/s)$ ............ $\times 1 0^{- 21}$ J.

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photoelectric Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy and Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Terms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Proportionality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought