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Physics-

General

Easy

Question

The ionisation energy of $Li to the power of 2 plus end exponent$ atom in ground state is :

13.6 x 9 eV
13.6J
13.6 erg
13.6x 10¹⁹-J

The correct answer is: 13.6 x 9 eV

Related Questions to study

A cylindrical solid of length L and radius a is having varying resistivity given by r = r0 x where r0 is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by :

A cylindrical solid of length L and radius a is having varying resistivity given by r = r0 x where r0 is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by :

physics-General

A charged oil drop falls with terminal velocity V₀ in the absence of electric field An electric field E keeps it stationary The drop acquires additional charge q and starts moving upwards with velocity V₀ The initial charge on the drop was

A charged oil drop falls with terminal velocity V₀ in the absence of electric field An electric field E keeps it stationary The drop acquires additional charge q and starts moving upwards with velocity V₀ The initial charge on the drop was

physics-General

if $ell parallel to m$ find the value of x in given figure.

if $ell parallel to m$ find the value of x in given figure.

parallel

The number of positive integral solutions of the inequation $fraction numerator x squared left parenthesis 3 x minus 4 right parenthesis cubed left parenthesis x minus 2 right parenthesis to the power of 4 over denominator left parenthesis x minus 5 right parenthesis to the power of 5 left parenthesis 2 x minus 7 right parenthesis to the power of 6 end fraction less or equal than 0$ is –

The number of positive integral solutions of the inequation $fraction numerator x squared left parenthesis 3 x minus 4 right parenthesis cubed left parenthesis x minus 2 right parenthesis to the power of 4 over denominator left parenthesis x minus 5 right parenthesis to the power of 5 left parenthesis 2 x minus 7 right parenthesis to the power of 6 end fraction less or equal than 0$ is –

Set of values of x satisfying the inequality $fraction numerator left parenthesis x minus 3 right parenthesis squared left parenthesis 2 x plus 5 right parenthesis squared left parenthesis x minus 7 right parenthesis over denominator open parentheses x squared plus x plus 1 close parentheses left parenthesis 3 x plus 6 right parenthesis squared end fraction less or equal than 0$ is $left square bracket a comma b right parenthesis union left parenthesis b comma c right square bracket$ then 2a + b + c is equal to

Set of values of x satisfying the inequality $fraction numerator left parenthesis x minus 3 right parenthesis squared left parenthesis 2 x plus 5 right parenthesis squared left parenthesis x minus 7 right parenthesis over denominator open parentheses x squared plus x plus 1 close parentheses left parenthesis 3 x plus 6 right parenthesis squared end fraction less or equal than 0$ is $left square bracket a comma b right parenthesis union left parenthesis b comma c right square bracket$ then 2a + b + c is equal to

Number of integral values of x for which the inequality log₁₀ $open parentheses fraction numerator 2 x minus 2007 over denominator x plus 1 end fraction close parentheses$ $less or equal than$ 0 holds true, is

Number of integral values of x for which the inequality log₁₀ $open parentheses fraction numerator 2 x minus 2007 over denominator x plus 1 end fraction close parentheses$ $less or equal than$ 0 holds true, is

parallel

$x to the power of log subscript 5 invisible function application x end exponent greater than 5$ implies

$x to the power of log subscript 5 invisible function application x end exponent greater than 5$ implies

log4 (2x2 + x + 1) – log² (2x – 1) $less or equal than$ – tan $fraction numerator 7 pi over denominator 4 end fraction$

log4 (2x2 + x + 1) – log² (2x – 1) $less or equal than$ – tan $fraction numerator 7 pi over denominator 4 end fraction$

The solution set of the inequation log1/3 (x² + x + 1) + 1 > 0 is

The solution set of the inequation log1/3 (x² + x + 1) + 1 > 0 is

parallel

If $open parentheses a to the power of log subscript b end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent$ –5 $x to the power of log subscript b end subscript invisible function application a end exponent$ + 6 = 0 where a > 0, b > 0 & ab $not equal to$ 1. Then the value of x is equal to

If $open parentheses a to the power of log subscript b end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent$ –5 $x to the power of log subscript b end subscript invisible function application a end exponent$ + 6 = 0 where a > 0, b > 0 & ab $not equal to$ 1. Then the value of x is equal to

No. of ordered pair satisfying simultaneously the system of equation $2 to the power of square root of x end exponent$ . $2 to the power of square root of y end exponent$ = 256 & log10 $square root of x y end root$ – log10 1.5 = 1 is.

No. of ordered pair satisfying simultaneously the system of equation $2 to the power of square root of x end exponent$ . $2 to the power of square root of y end exponent$ = 256 & log10 $square root of x y end root$ – log10 1.5 = 1 is.

If $x to the power of left square bracket log subscript 3 end subscript invisible function application x to the power of 2 end exponent plus left parenthesis log subscript 3 end subscript invisible function application x right parenthesis to the power of 2 end exponent minus 10 right square bracket end exponent$ = 1/x², then x =

If $x to the power of left square bracket log subscript 3 end subscript invisible function application x to the power of 2 end exponent plus left parenthesis log subscript 3 end subscript invisible function application x right parenthesis to the power of 2 end exponent minus 10 right square bracket end exponent$ = 1/x², then x =

parallel

If x_n > x_n–1>...> x₂ > x₁ > 1 then the value of $log subscript straight x subscript 1 end subscript invisible function application log subscript straight x subscript 2 end subscript invisible function application log subscript straight x subscript 3 end subscript invisible function application horizontal ellipsis log subscript straight x subscript straight n end subscript invisible function application x subscript n$ $blank to the power of x subscript n minus 1 end subscript superscript up right diagonal ellipsis to the power of x subscript 1 end exponent end superscript end exponent$ is equal to-

If x_n > x_n–1>...> x₂ > x₁ > 1 then the value of $log subscript straight x subscript 1 end subscript invisible function application log subscript straight x subscript 2 end subscript invisible function application log subscript straight x subscript 3 end subscript invisible function application horizontal ellipsis log subscript straight x subscript straight n end subscript invisible function application x subscript n$ $blank to the power of x subscript n minus 1 end subscript superscript up right diagonal ellipsis to the power of x subscript 1 end exponent end superscript end exponent$ is equal to-

The expression logp where $p greater or equal than 2 comma p element of N semicolon n element of N$ when simplified is.

The expression logp where $p greater or equal than 2 comma p element of N semicolon n element of N$ when simplified is.

Let N= $open parentheses open parentheses square root of 7 close parentheses to the power of fraction numerator 2 over denominator log subscript 25 end subscript invisible function application 7 end fraction end exponent minus 12 5 to the power of log subscript 25 end subscript invisible function application 6 end exponent close parentheses$ Then log₂N has the value –

Let N= $open parentheses open parentheses square root of 7 close parentheses to the power of fraction numerator 2 over denominator log subscript 25 end subscript invisible function application 7 end fraction end exponent minus 12 5 to the power of log subscript 25 end subscript invisible function application 6 end exponent close parentheses$ Then log₂N has the value –

parallel

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