Question

In a new system of units the fundamental quantities are planks constant $\left(h\right)$ speed of light $\left(c\right)$ and time $\left(T\right)$. If the dimensions of Rydberg's constant is $h^{-x}{c}^{-y}{T}^{-z}$, then what is the value of $x+y+z$ ?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

Planck's constant, [h]= $\left[M^1{L}^2{T}^{-1}\right]$

Speed of light $\left[c\right]=\left[M^0{L}^1{T}^{-1}\right]$

Time, $\left[t\right]=\left[T\right]$

Rydberg's constant, $\left[R\right]=\left[M^0{L}^{-1}{T}^0\right]$

$\left[R\right]=\left[h{]}^{-x}\right[c{]}^{-y}[T{]}^{-z}$

$\left[M^0{L}^{-1}{T}^0\right]=\left[M^1{L}^2{T}^{-1}{]}^{-x}\right[M^0{L}^1{T}^{-1}{]}^{-y}[T{]}^{-z}$

$\therefore \left[L^{-1}\right]=\left[M^{-x}\right]\left[L^{-y-2x}\right]\left[T^{x+y-z}\right]$

$\therefore -x=0\Rightarrow x=0$

$-y-2x=-1$

$\Rightarrow -y-0=-1\Rightarrow y=1$

$\therefore x+y-z\Rightarrow 0+1-z=0$

$\Rightarrow z=1$

Now, $x+y+z=0+1+1=2$