JEE Main 2019 (Online) 11th January Evening Slot | Structure of Atom Question 151 | Chemistry

The de Broglie wavelength ($$\lambda $$) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v₀ is threshold frequency] :

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{3 \over 2}}}}}$$

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 4}}}}}$$

$$\lambda \,\infty \,{1 \over {\left( {v - {v_0}} \right)}}$$

$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose?

[R_H = 1 $$ \times $$ 10⁵ cm^–1, h = 6.6 $$ \times $$ 10^–34 Js, c = 3 $$ \times $$ 10⁸ ms^–1]

Balmer, $$\infty $$ $$ \to $$ 2

Paschen, 5 $$ \to $$ 3

Paschen, $$\infty $$ $$ \to $$ 3

Lyman, $$\infty $$ $$ \to $$ 1

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He⁺ ion in eV is :

$$-$$ 6.04

$$-$$ 54.4

$$-$$ 27.2

$$-$$ 3.4