NEET Chemistry Solved Question Paper 2006

E = $\frac{\mathrm{hc}}{\mathrm{\lambda }}$ or E $\propto$ $\frac{1}{\mathrm{\lambda }}$

Thus a decrease in wave length represents an increase in energy For n₄ $\to$ n₁ transition .These are greater the. energy difference and lesser will be wavelength .Thus , for n₄ $\to$ n₁  transition has minimum wavelength .

According to Boyle's law , "for a given mass of a gas , at constant temperature the volume of a gas is inversely proportional to its pressure" .

V $\propto$ $\frac{1}{\mathrm{P}}$

or PV = constant

on differentiating the equation

d(PV) = d        (constant)

= PdV + VdP =0

= VdP = - PdV

= $\frac{\mathrm{dP}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{V}}$

$\underset{(\mathrm{strong} \mathrm{acid})}{\mathrm{HCl}}$ + $\underset{(\mathrm{strong} \mathrm{base})}{\mathrm{KOH}}$ $\stackrel{\mathrm{neutralisation}}{\rightleftharpoons }$ KCl + H₂O

In this reaction , HCl is the strong acid and KOH is the base and it has been found that the heat of neutralisation of a strong acid with a strong base is always constant i.e. , 57.3 kJ .

An ionic bond is formed between an electropositive metal and a electronegative non metal .The electropositive metal converts into a cation , while the electronegative non metal converts into a anion .

Thus the formation ofionic bond is favoured by

(I) Low ionisation potential of metal

(II) Greater valtie of electron affinity of non metal

(III) Higher value of lattice energy of the resulting ionic compound .

Mg(OH)₂ $\rightleftharpoons$ Mg²⁺ + 2OH^-

the solubility product K_sp of

Mg(OH)₂ = [Mg²⁺][OH^-]²

1 x 10^-12 = 0.01 [OH^-]²

[OH^-]² = 1 x 10^-10

[OH^-] = 10^-5

We know

[H⁺][OH^-] = 10­^-14

[H⁺][10^-5] = 10^-14

[H] = 10^-14/10^-5 = 10^-9

pH = - log[H⁺] = - log 10^-9 = 9

An atomic orbital is a three dimensional region of definite shape around the nucleus where the probability of finding the electron is maximum .Therefore , an atom has a both characteristic energy and a characteristic shape .

According to aufbau principle , "sub-shells are filled with electrons in the increasing order of their energies , "i.e. , sub-shell of lower energy will be filled first with electrons .

Thus correct order is -

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d ... etc

A metal surface is exposed to solar radiations ,the emitted electrons have energy less than maximum value of energy depending upon freauency of incident radiations .

We know ,

N_t = N_o x ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

where N_t = amount left after expiry of 'n' half lives

N_o = initial amount

n = number of half lives elapsed

$\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_{\mathrm{o}}}$ = ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

$\frac{1}{64}$ = ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

${\left(\frac{1}{2}\right)}^6$ = ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

n = 6

Time taken (T) = t_1/2 x n = 2 x 6 = 12 h

According to Heisenberg's uncertainty principle

$∆$x ^. $∆$p $⩾$ $\frac{\mathrm{h}}{4\mathrm{\pi }}$    ($∆$p = m$∆\mathrm{\nu }$)

$∆$x ^. m$∆$v $⩾$ $\frac{\mathrm{h}}{4\mathrm{\pi }}$

$∆$x = $\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$