Question

The ${\lambda }_{min}$ of a continuous X-Ray spectrum is 0.414 $A^0$ , when it operates at a p.d. of 30 KV. The value of plancks constant, h, is (given e $=1.6\times {10}^{-19}$ C, $c=3\times {10}^8$ m/s)

• A. $6.624\times {10}^{-34}Js$
• B. $6.524\times {10}^{-34}Js$
• C. $6.674\times {10}^{-34}Js$
• D. $6.724\times {10}^{-34}Js$

Answer 1

The correct option is A $6.624\times {10}^{-34}Js$

${\lambda }_{cutoff}=\frac{he}{ev}$

$h=\frac{0.414\times {10}^{-4}\times 1.6\times {10}^{-19}\times 30\times {10}^3}{3\times {10}^8}$

$=6.624\times {10}^{-34}Is$

So, the answer is option (A).