The graph that correctly represents the relat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Moseley's law,the frequency $
u$ of a characteristic $X$-ray is related to the atomic number $Z$ of the target material by the equat

Vedclass · Accepted Answer

Vedclass · Answer

(C) According to Moseley's law,the frequency $
u$ of a characteristic $X$-ray is related to the atomic number $Z$ of the target material by the equation:$\sqrt{
u} = a(Z - b)$where $a$ and $b$ are constants.Squaring both sides,we get:$
u = a^2(Z - b)^2$This equation is of the form $
u = k(Z - b)^2$,which represents a parabola opening upwards with its vertex at $(b, 0)$ on the $Z$-axis.Comparing this with the given options,the graph that represents a parabolic relationship between $
u$ and $Z$ is graph $C$.

The graph that correctly represents the relation of frequency $\nu$ of a particular characteristic $X$-ray with the atomic number $Z$ of the material is

Similar Questions

The $K_{\alpha}$ $X$-ray emission from tungsten occurs at $\lambda = 0.021 \ nm$. The energy difference between the $K$ and $L$ levels in the atom is:

The wavelength of the ${K_\alpha }$ line in copper is $1.54 \mathring A$. The ionisation energy of the $K$ electron in copper in Joules is:

The wavelength of the $K_{\alpha}$ line for an element with atomic number $57$ is $\lambda$. What will be the wavelength of the $K_{\alpha}$ line for an element with atomic number $29$?

$X$-rays of wavelength $0.1 \ \mathring A$ are allowed to fall on a metal and get scattered. The wavelength of the scattered radiation is $0.111 \ \mathring A$. If $h = 6.624 \times 10^{-34} \ J \cdot s$ and $m_0 = 9 \times 10^{-31} \ kg$,then the direction of the scattered photons will be:

When the potential difference applied to an $X$-ray tube is increased, the resulting emitted radiation shows:

Vedclass Test Series

Exam Paper Generator

Online Exam Module