Match List I with List II: List I Lis | vedclass

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

Question

(B) The dimensional formulas are calculated as follows:$(A)$ Planck's constant $(h)$: From $E = h
u$,we have $h = E / 
u$. The dimensions of energy

Vedclass · Accepted Answer

$A-III, B-IV, C-I, D-II$

Vedclass · Answer

(B) The dimensional formulas are calculated as follows:$(A)$ Planck's constant $(h)$: From $E = h
u$,we have $h = E / 
u$. The dimensions of energy $E$ are $[M^1 L^2 T^{-2}]$ and frequency $
u$ are $[T^{-1}]$. Thus,$[h] = [M^1 L^2 T^{-2}] / [T^{-1}] = [M^1 L^2 T^{-1}]$. This matches $III$.$(B)$ Stopping potential $(V_s)$: From $E = qV$,we have $V = E / q$. The dimensions of energy $E$ are $[M^1 L^2 T^{-2}]$ and charge $q$ are $[A^1 T^1]$. Thus,$[V_s] = [M^1 L^2 T^{-2}] / [A^1 T^1] = [M^1 L^2 T^{-3} A^{-1}]$. This matches $IV$.$(C)$ Work function $(\phi)$: Work function is a form of energy. Thus,$[\phi] = [M^1 L^2 T^{-2}]$. This matches $I$.$(D)$ Momentum $(p)$: Momentum $p = mv$. The dimensions are $[M^1] 	imes [L^1 T^{-1}] = [M^1 L^1 T^{-1}]$. This matches $II$.Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

List $I$	List $II$
$A$. Planck's constant $(h)$	$I$. $[M^1 L^2 T^{-2}]$
$B$. Stopping potential $(V_s)$	$II$. $[M^1 L^1 T^{-1}]$
$C$. Work function $(\phi)$	$III$. $[M^1 L^2 T^{-1}]$
$D$. Momentum $(p)$	$IV$. $[M^1 L^2 T^{-3} A^{-1}]$

Match List $I$ with List $II$:

List $I$ List $II$

$A$. Planck's constant $(h)$ $I$. $[M^1 L^2 T^{-2}]$

$B$. Stopping potential $(V_s)$ $II$. $[M^1 L^1 T^{-1}]$

$C$. Work function $(\phi)$ $III$. $[M^1 L^2 T^{-1}]$

$D$. Momentum $(p)$ $IV$. $[M^1 L^2 T^{-3} A^{-1}]$

Similar Questions

The graph shows the stopping potential as a function of the frequency of incident radiation for two different surfaces $A$ and $B$. The work function of surface $A$ is ......

In the following arrangement $y = 1.0\ mm$,$d = 0.24\ mm$ and $D = 1.2\ m$. The work function of the material of the emitter is $2.2\ eV$. The stopping potential $V_0$ needed to stop the photocurrent will be .............. $V$.

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the wavelength of the electron to that of the photon is: ($c$ being the velocity of light)

Two electrons are moving with the same speed $v$. One electron enters a region of a uniform electric field,while the other enters a region of a uniform magnetic field. If the de-Broglie wavelengths of the two electrons after some time are ${\lambda _1}$ and ${\lambda _2}$ respectively,then:

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength $\lambda_{1}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match List $I$ with List $II$: List $I$ List $II$ $A$. Planck's constant $(h)$ $I$. $[M^1 L^2 T^{-2}]$ $B$. Stopping potential $(V_s)$ $II$. $[M^1 L^1 T^{-1}]$ $C$. Work function $(\phi)$ $III$. $[M^1 L^2 T^{-1}]$ $D$. Momentum $(p)$ $IV$. $[M^1 L^2 T^{-3} A^{-1}]$