Question

When photons of energy $4.25\text{eV}$ strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy $T_A\text{eV}$ and de-Broglie wave length ${\lambda }_A$. The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy $4.70\text{eV}$ is $T_B=(T_A–1.50)eV.$ If the de-Broglie wave length of these photo electrons is ${\lambda }_B=2{\lambda }_A,$ then:

• A.
  The work function of A is $2.25\text{eV}$
• B.
  The work function of B is $4.20\text{eV}$
• C.
  $T_A=2.00\text{eV}$
• D.
  $T_B=2.75\text{eV}$

Answer 1

Answer: (A), (B), (C)

$T_A=2.00\text{eV}$
For metal $\text{A}:$

Maximum kinetic energy , $T_A=h\nu -W_A$

From the data given in the question,

$T_A=4.25-W_A........\left(1\right)$

But for a particle of mass $m$, kinetic energy is given by $K=\frac{p^2}{2m}$

Since, electron is infinitesimally small, the de-broglie wave associated with the motion of electron is given by ${\lambda }_e=\frac{h}{p_e}$

Thus, $T_A=\frac{h^2}{2m_e{\lambda }_A^2}......\left(2\right)$

For metal $\text{B}:$

Maximum kinetic energy , $T_B=h\nu -W_B$

From the data given in the question,

$T_B=4.70-W_B.....\left(3\right)$

Similarly, using the de-broglie wavelength we can write that,

$T_B=\frac{h^2}{2m_e{\lambda }_B^2}.....\left(4\right)$

But, $T_B=T_A-1.50$

Substituting this in $\left(3\right)$, $T_A=6.2-W_B....\left(5\right)$

From $\left(5\right)$ and $\left(1\right)$ , $W_B-W_A=1.95$

From $\left(2\right)$ and $\left(4\right)$ , $\frac{T_A}{T_B}={\left(\frac{{\lambda }_B}{{\lambda }_A}\right)}^2$

But, given that ${\lambda }_B=2{\lambda }_A$

$\therefore \frac{T_A}{T_A-1.5}=4\Rightarrow T_A=2\text{eV}$

Therefore , $T_B=2-1.50=0.5\text{eV}$

$W_A=2.25\text{eV}$ and $W_B=2.25+1.95=4.20\text{eV}$

Hence, options (a) , (b) and (c) are the correct answers.