Question

If U.V. light of wavelengths 800 $A^0$ and 700 $A^0$ can liberate electrons with kinetic energies of 1.8eV and 4 eV respectively from hydrogen atom in ground state, then the value of Planck's constant is

• A. $6.57\times {10}^{-34}Js$
• B. $6.63\times {10}^{-34}Js$
• C. $6.66\times {10}^{-34}Js$
• D. $6.77\times {10}^{-34}Js$

Answer 1

Answer: (A)

$6.57\times {10}^{-34}Js$

From first eqn of photoelectric effect,
$\Rightarrow (\frac{hC}{8}\times {10}^8)-W=1.8$ ---------------(I)
From second eqn of photoelectric effect,
$\Rightarrow \frac{hC\times {10}^8}{7}-W=4$ -----------------(II)
Solving (I) and (II) simultaneously we get,
$h=4.10625\times {10}^{-15}eVs$
Converting this $h$ in terms of $Js$ we get,
$h=4.10625\times 1.6\times {10}^{-19-15}$
$h=6.57\times {10}^{-34}Js$

So, the answer is option (A).