Atomic Structure Result Question 8-1
8. The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
(2019 Main, 11 Jan II)
(a) $\lambda \propto \dfrac{1}{\left(v-v_0\right)^{\dfrac{1}{4}}}$
(b) $\lambda \propto \dfrac{1}{\left(v-v_0\right)^{\dfrac{3}{2}}}$
(c) $\lambda \propto \dfrac{1}{\left(v-v_0\right)}$
(d) $\lambda \propto \dfrac{1}{\left(v-v_0\right)^{\dfrac{1}{2}}}$
Show Answer
Answer:
Correct Answer: 8. ( d )
Solution:
- de-Broglie wavelength $(\lambda)$ for electron is given by
$\lambda=\dfrac{h}{\sqrt{2 m \mathrm{~K} \cdot \mathrm{E}}}\quad$ …..(i)
Also, according to photoelectric effect
$\mathrm{KE}=h \nu-h v_0$
On substituting the value of $KE$ in Eq (i), we get
$\begin{aligned} & \lambda=\dfrac{h}{\sqrt{2 m \times\left(h v-h v_0\right)}} \\ & \lambda \propto \dfrac{1}{\left(v-v_0\right)^{1 / 2}} \end{aligned}$
BETA
